C-Michigan State University

CSE 251 Dr. Charles B. OwenProgramming in C1

Getting Started

Sakti Pramanik(Instructor)

2148 E. B.

office hours

Thu 2-4

Soroor Soltani(TA and grading)

Classroom: 3353 Engineering Building


Structure of the Course

Get a quick intro to C programming

Learn enough Unix to use it

Get some experience with

C-oriented tools.Get some experience

writing code


What is C?

C is a rather old programming language (1972, Richie, Bell Labs)

Originally designed as a systems software platform (OS and the

like)

Procedural, block oriented language

(no object-oriented programming)Dennis RitchieInventor of the

C Programming Language


Why learn C?• Small, extensible language. Progenitor of many

languages.• Many applications and much support due to its

age and general use• Many tools written to support C development.• Close to the hardware, programmer manages

memory• Common embedded systems language• Can be fast, efficient (most cited reason)


Disadvantages of C?• Flexible language allows programmers to commit

many sins without warning• Hard to debug, fix code• Speed depends on programmer as much as the

language• Managing memory can be dangerous, difficult,

painful• Lacks modern features (OOP, exceptions, etc.)


Course Structure


Course will be Lecture/Step

This is a 1 credit course, meets only 15 times

Lecture/Step approach

pend some time on a lecture topic.

pend some time doing “hands-on”work.

ach lecture/step will have some exercise you will have to turn in at the

end


Grading

3 projects, each for 15% of the grade, total 45%

13-15 Step assignments, exercises, total 55%

That’s it. No exams, no quizzes


Attendance is important

We drop the lowest step assignment grade!

Should go without saying that in a course like this attendance is important.


Step exercises

• These can be collaborative, done with discussion and help from anyone during lab. Talk to people, ask questions.

• Lab time is a time to figure stuff out


Projects are individual

• No collaboration on projects• They will be checked by cheat check• Just don’t do it. Do your own work

• No makeups! There are only 3 and I’ll give plenty of notice. Get it in on time.


Get Started, booting


Rebooting to linux

• Machines are split to run either linux (a unixvariant) or Windows

• You can reboot to run “diskless” and bring any machine up in linux.

• Let’s do that now!


Booting

If your machine says “press control-alt-delete to log in”, do so, but use the red button in the lower right corner to shut the machine down. Then restart, selecting “diskless” when prompted.

Before logging in, use the Session menu to select GNOME as your session.

Once logged in, select Applications/Internet/IceweaselWeb Browser.

Go to http://www.cse.msu.edu/~cse251

Initial password is your PID (starts with A)

0


Bringing up…

Bring up a terminal Window

Applications/Accessories/Terminal

Type:

cal 2011


Directories


What is a directory/folder?

Be it Windows, Linux or OS X, all OS’s maintain a

directory structure.

A directory is a container of files or other directories

These directories are arranged in a hierarchy or

tree


Directory StructureHas a root node, with branch nodes, ends in leaf nodes

The directory structure is a tree

Each directory can hold files and ‘point to’ parent and children directories

Root

Branches

Leaves

/user /bin

/

/cse251

/exercises

hello.c

a.out

/john

a.out

ls

myhello.c


File PathA path to a file is a path through the hierarchy to the node that contains a file or directory

/user/cse251/exercises/hello.c myhello.c

/user /bin

/

/cse251

/exercises

hello.c

a.out

/john

a.out

ls

Path is from root node /, to user directory, to cse251 directory, to exercises directory, where the file hello.c resides.


The current directory

As you type in the command shell, you are presently “in” one current directory

Many commands are available to navigate by changing your current directory


Unix command caution(s)

Names are short. This is to avoid typing. Get used to it

Most commands take “switches”, parameters which modify the behavior of the command. Usually preceded by a ‘-’

All commands, all switches, all paths, all filenames are case sensitive


Some Unix Commands

pwd – Displays the current working directory

ls – Lists the contents of the current directory

cd – Changes the current directory

mkdir – Creates a new directory

1


Path String

a valid path string is a series of directories (separated by ‘/’) which indicates a valid path in the directory structure

‘/user/cse251/exercises/hello.c’ is a valid path string but ‘/cse251/hello.c’ is not


Three special directory names• ‘.’ is shortcut for current directory you are in

• ‘./a.out’ is the same as ‘/user/cse251/exercises/a.out’ if you’re currently in /user/cse251/exercises directory

• ‘..’ is shortcut for the name of the parent directory of the current directory you are in• ‘../a.out’ is the same as ‘/user/cse251/a.out’ if you’re

currently in /user/cse251/exercises directory

• ‘~’ is a shortcut for your home directory• ‘~/myhello.c’ is the same as ‘/user/john/myhello.c’ if your

login name is john

2


mv (changes a file)mv is the move command. Rename a file in place or move to a new directory

• mv file.c newFile.c• rename file.c to newFile.c in current directory

• mv file.c /user/ptan/• move file.c in the current directory to /user/ptan. The target

directory must already exist

• mv ~/file.c ./newFile.c• move file.c in home directory to the current directory with the

name newFile.c


cp, copy a file

• cp file.c newFile.c• Create a new file called newFile.c in current directory

from the file file.c, which is also in current directory• cp ../file.c ~/programs

• cp file.c in parent directory to the sub-directory programs under the home directory


rm, remove a file

Cannot be undone, be careful• rm /user/cse251/exercises/a.out

• remove the file called a.out• rm ./file.c

• remove file.c in the current directory• rm –i ./file.c

• interactive, are you sure?

3


man

man pages exist on Unix, providing documentation (a lot of documentation) on each command

• man ls• man page on the ls command

• man -k graphics or apropos graphics• every man page that has “graphics” as a word

• man -S 2 mount• man page for section 2 of mount


Better man pages

• Almost every distribution of Linux provides a better way to view man pages

• On Debian/Ubuntu, the program yelp (the help program) will show man pages

• yelp, then man ls• show man page on ls in a nice way

• yelp runs using the “lifesaver” icon. Go to Advanced options, man pages

4


What’s the shell

• the shell is the program that interacts with you through the terminal window

• there are many, and you can change it easily• csh• ksh• tcsh• bash

• by default, you are using tcsh


Prompt

Default prompt looks like:• <47 nelson:/usr/include >

• 47, which command in the history this is• nelson, the name of the machine you are on• /usr/include, current directory

can be configured differently• [19:19][31][cse251@nelson]~• >

• two lines: time, history, who@machine, path


Job control

Only one job runs in foreground (sending text to console), but many can run in background

Foreground job controls the console

• emacs myFile.c &• & mean run emacs in background. Console remains

responsive to commands

• jobs• lists all jobs running in background. Numbers can be used to

control job


ps: List of processes

• Typically, each job that is executing is called a process• Sometimes, a job can produce (fork) multiple

processes

• Use “ps” command to list all your current processes (including those that were suspended or running in background)• It also shows the background processes (including

the shell program you’re using)

5


Command completion

Type a partial command, followed by tab. it will complete as much as it can


Compiling


compile to make an executable

• In C, you cannot directly execute the source code. You go through a process called compilation

• Compilation makes sure you “followed the rules of C” (even if you did something stupid), and makes an executable

• Compilation errors are rule “mistakes”. A compilation error means no executable

• The executable file is the thing you run


gcc: gnu compiler

• There are others, but this is a good one and it is free (and on all Linux distributions, can be added to the MacOS and runs under Cygwin on windows)

• Has just around a zillion switches. Can be pretty impressive if you read the man page.


Example gcc usage

• gcc file.c• if it compiles, makes an executable called a.out in the

same directory as file.c• can run a.out by typing ./a.out (meaning the a.out in

the current directory)• gcc -o myExec file.c

• make the executable called myExec


Compilation errors

• You didn’t follow the rules• You have to fix the source code before the

compiler can make an executable• Errors can be cryptic, but at the very least they

list the line number where it went wrong• Doesn’t prevent run-time errors


Editors


Many Editors

Examples: • gedit• vi • emacs• pico• kate

6

I’m going to suggest using gedit in this course.

Introducing C

1. Please get logged in.

2 Open a new terminal window Applications/Accessories/Terminal2. Open a new terminal window Applications/Accessories/Terminal

3. Open a web browser Applications/Internet/Iceweasel Web Browser

4. Go to http://www.cse.msu.edu/~cse251

5. Open Step 2: Introducing C


Hello World

#include <stdio.h>

/*/* This is my first program in CSE 251*/

int main(){{printf("Hello, and welcome to CSE 251!!!\n");

}


Hello World

#include <stdio.h>

/*

#include statement

I l d th d i t/* This is my first program in CSE 251*/

Includes other source code into your program. One use is to tell your program about libraries you

int main(){

your program about libraries you are using. We are using the Standard Input/Output library {

printf("Hello, and welcome to CSE 251!!!\n");}

right now: stdio. The file stdio.h is the header that defines the S d d I/O LibStandard I/O Library.


Hello World

#include <stdio.h>

/*

Comment

/* This is my first program in CSE 251*/

int main(){

A Comment is a note to yourself. In C we start a comment with /* and end it with */ Everything in{

printf("Hello, and welcome to CSE 251!!!\n");}

and end it with */. Everything in between is ignored.

Programs can be cryptic and hard to read. You need to remind yourself what you were thinking when you wrote something.


That’s what comments are for.

Hello World The main function

#include <stdio.h>

/*main() is the starting point for a C

I ’ h h/* This is my first program in CSE 251*/

program. It’s where the computer begins execution. Note the syntax: the code is contained in curly

int main(){

the code is contained in curly braces.

{printf("Hello, and welcome to CSE 251!!!\n");

}


Hello World A printf function call

#include <stdio.h>

/*/* This is my first program in CSE 251*/

printf() is a library function that prints. It’s the main way we will get output from our C programs

int main(){

get output from our C programs.

{printf("Hello, and welcome to CSE 251!!!\n");

}


printf statement in more detailA f i ll ( h d i )A function call (what we are doing)Always a name followed by (, optional arguments, ), and ;.

A string – Content between " and " is our way of telling the computer about

{printf("Hello and welcome to CSE 251!!!\n");

our way of telling the computer about text we want to print.

printf( Hello, and welcome to CSE 251!!!\n );}

\n means “newline”.

C statements end with a ;

CSE 251 Dr. Charles B. OwenProgramming in C7 1

Escape characters

Characters that are hard to express:• \n newline• \t tab• \’ print a single quote\ p g q• \\ print a backslash• many othersmany others


Variable Declarations

Here is how C deals with memory:• Imagine the system memory as a nice, flat stretch of beach• You want a variable, you need to dig a hole in the sand and dump the value in

• How big a hole?How big a hole?

The holes in C are very specialized. Once dug, they can only hold one type of thing Our holethey can only hold one type of thing. Our hole might hold integers, floating point values, or characters, for example.


Declare variable before useDeclare variable before use

• When you declare a variable, you are telling the y y gcompiler the kind of value the variable may hold (its type)

• You cannot change the type of value a variable can hold once declared (well, pretty much anyway)

• In fact, everything needs a type in C and it must be declared before use!


Common types, “regular” C

• int : an integer, usually 4 bytes• float: float, usually 4 bytes• double : float, usually 8 bytes• char : single char, value in single quotesg , g q


Must declare before use

• Every variable must be declared before it can be used (its type must be indicated)

• Syntax:<variable_type> <variable_name> [ =<initial_value> ];

(optional)

• Example: int length;


Must declare before use

• Every variable must be declared before it can be used (its type must be indicated)

• Syntax:<variable_type> <variable_name> [ =<initial_value> ] ;

(optional)

• Example: double width = 10;


Rules for Variable Names

• Must begin with a letter• Any combination of letters, digits andAny combination of letters, digits and underscore

• Up to 31 characters long• Up to 31 characters long• Cannot match with a C keyword

– E.g., int int; int long;


Some C Keywordsyauto double int structbreak else long switchbreak else long switchcase enum register typedefh t t ichar extern return unionconst float short unsigned

fcontinue for signed voiddefault goto sizeof volatiledo if static while


Some Variable Declarations

int numLots = 7;double width1;double width2;float carHP = 420.7;i bint a, b, c;


printf: a new feature

You can use printf to output variables as well as strings. Put a “descriptor” in the string you print:

int numLots = 10;double totalArea = 100;double totalArea 00;printf(“There are %d lots\n”, numLots);printf(“The %d lots have a total area of %f\n”, numLots, totalArea);

Follow the string with variables.Each descriptor is replaced with a variable’s value.p p


Many descriptors

• %s : string• %d : decimal (integer)• %d : decimal (integer)• %e : floating point exponent• %f : floating point decimal• %u : unsigned integer• %u : unsigned integer• and others


Full Format stringg

• The format string contains a set of format descriptors that describe how an object is to be printedthat describe how an object is to be printed

% -#0 12 .4 h d ConversionType

startifi ti d h

sizemodifier

Type (d for decimal)

specification

Flags

Width

Precision

(h for short int)


Examplesf(“ f\ ” )• printf(“%f\n”,M_PI);

– 3.141593

i f(“% 4f\ ”M PI)• printf(“%.4f\n”,M_PI);– 3.1416 (4 decimal points of precision, with rounding)

i tf(“%10 2f\ ”M PI)• printf(“%10.2f\n”,M_PI);– 3.14 (10 spaces total including the number and the decimal point)decimal point)

• printf(“%‐10.2f is PI\n”,M_PI);– 3.14 is PI (10 spaces, but left justified)( p , j )


Examplesi t L t 10int numLots = 10;double totalArea = 100.7883;float current = 22.7;

printf("There are %d lots with a total area of %f\n", numLots, totalArea);/* Output: There are 10 lots with a total area of 100.788300 */

printf("The area is %6.2f\n", totalArea);/* Output: The area is 100.79 */

printf("The meter reads %.1f amps\n", current);/* Output: The meter reads 22.7 amps */


Expression

An expression is written in pretty standard mathematical notation:

wins + losseswidth * heightwidth heightarea / widthx * y * z – qM PI * 2M_PI * 2

printf(“We played %d games\n”, wins + losses);

We can put expressions where we have been putting variables.


Types determine results

• For integers: +,‐,*,/ all yield integers. Thus division can lead to truncation (2/3 has value 0). % gives the remainder

• For floats: +,‐,*,/ all work as advertised. No remainder.


Mixed computation

• As with most languages, C expects to work with like types. 1 + 1, 3.14 + 4.56

• When mixing, errors are common except where C can “help”

• It will promote a value to a more “detailed” type when required

• 1 + 3.14 yields a float (1 promoted to 1.0)


Assignment

An assignment puts a value in a variable. It is of the form:

variable = expression ;variable = expression ;

Do everything on the right hand side, get a value.Dump the value into the variable.

double area, circumference;, ;area = width * height;circumference = radius * 2 * M_PI;

Declare first, assign later!Don’t forget the semicolon.


Expression and Assignment Examples

double volume, diameter, hypot;int games;

volume = radius * radius * M_PI;diameter = radius * 2;diameter = radius 2;games = wins + losses;hypot = sqrt(near * near + pow(far, 2));yp q p

To use M_PI or math functions: #include <math.h> and compile with –lm switch.


scanf

printf("Please enter the yards of pipe used: ");scanf(”%f",&yardsOfPipe);

Scanf is an input routine• Useful for reading in string input and doing• Useful for reading in string input and doing conversion to the correct type, all at once

• Syntax is “kind of like” printfSyntax is kind of like printf• Note the use of the & operator!!!


Basic form

• To understand input, it is probably better to start with an example.

scanf(“%d, %f”, &myInt, &myFloat); floatint

…is waiting for input of the exact form

floatint

25, 3.14159

Use %f for float and %lf for double (That’s the letter l not the number 1)Use %f for float and %lf for double. (That’s the letter l, not the number 1)


format string the same

• What is typed in the format string is the same as what the input expects, in this case:– a decimal number– a comma

fl i i b– a floating point number


The &

• Hard to explain at the moment, but any variable that gets read in needs that ampersand character

• It is the address of the variable• more on that later


scanf examples

printf("Input the radius: ");scanf("%lf", &radius);

printf("Input the height: ");f("%lf" &h i h )scanf("%lf", &height);

printf("Input the number of cylinders: ");printf( Input the number of cylinders: );scanf("%d", &numCylinders);


Flow Control and Booleans

1. Please get logged in.

2 Open a new terminal window Applications/Accessories/Terminal2. Open a new terminal window Applications/Accessories/Terminal

3. Open a web browser Applications/Internet/Iceweasel Web Browser

4. Go to http://www.cse.msu.edu/~cse251

5. Open Step 3: Flow Control and Booleans


Today: Flow Control and Booleans

Flow Controlif, switch

Boolean Logic<, <=, ==, >=, >, !=


rlc.c#include <stdio.h>#include <math.h>

/* /* Simple program to compute the resonant frequency of * an RLC circuit*/

int main(){

double l; /* Inductance in millihenrys */double c; /* Capacitance in microfarads */; / p /double omega; /* Resonance frequency in radians per second */double f; /* Resonance frequency in Hertz */

printf("Enter the inductance in millihenrys: ");p ( y );scanf("%lf", &l);

printf("Enter the capacitance in microfarads: ");scanf("%lf", &c);( , );

omega = 1.0 / sqrt((l / 1000) * (c / 1000000));f = omega / (2 * M_PI);printf("Resonant frequency: %.2f\n", f);


p ( q y , );}


/* /* Simple program to compute the resonant frequency of * an RLC circuit*/ Note the use of comments to tell: a) what

the program does, b) what some lines ofint main(){

double l; /* Inductance in millihenrys */double c; /* Capacitance in microfarads */

the program does, b) what some lines of code do, and c) what the variables are.

; / p /double omega; /* Resonance frequency in radians per second */double f; /* Resonance frequency in Hertz */



omega = 1.0 / sqrt((l / 1000) * (c / 1000000));f = omega / (2 * M_PI); /* Convert radians per sec to Hertz */printf("Resonant frequency: %.2f\n", f);


p ( q y , );}


/* /* Simple program to compute the resonant frequency of * an RLC circuit*/ Note the use of indentation

int main(){

double l; /* Inductance in millihenrys */double c; /* Capacitance in microfarads */; / p /double omega; /* Resonance frequency in radians per second */double f; /* Resonance frequency in Hertz */



omega = 1.0 / sqrt((l / 1000) * (c / 1000000));f = omega / (2 * M_PI);printf("Resonant frequency: %.2f\n", f);


p ( q y , );}


/* /* Simple program to compute the resonant frequency of * an RLC circuit*/ Indentation is necessary to

make your program readable!int main(){double l; /* Inductance in millihenrys */double c; /* Capacitance in microfarads */

make your program readable!

; / p /double omega; /* Resonance frequency in radians per second */double f; /* Resonance frequency in Hertz */printf("Enter the inductance in millihenrys: ");scanf("%lf", &l);( , );printf("Enter the capacitance in microfarads: ");scanf("%lf", &c);omega = 1.0 / sqrt((l / 1000) * (c / 1000000));f = omega / (2 * M PI);g ( _ );printf("Resonant frequency: %.2f\n", f);}


Sequential Execution

Statement 1

Statement 2

...

Statement n


Selective Execution

booleanexpression

true false

statement 1 statement 2statement 1 statement 2


if statements

The if statementif(age > 39)Fundamental means of flow controlHow we will make decisions

( g )printf(“You are so old!\n”);

Boolean expressionsThe actual determination of

the decision

age > 39c == 0l <= 0(age >= 18) && (age < 65)the decision (age > 8) && (age < 65)


Structure of an if statement

If expression1 is true, execute statement1.

Otherwise test to see if expression2 is true If

if(expression1)

Otherwise, test to see if expression2 is true. If so, execute statement2.

Otherwise, execute statement3.statement1;

else if(expression2) /* Optional */statement2;

Otherwise, execute statement3.

statement2;else /* Optional */

statement3;The expressions are boolean expressions thatThe expressions are boolean expressions that resolve to a true or a false.


Basic Boolean Expressions Some operators:

truefalse

< Less than<= Less than or equal to

age < 18divisor == 0size > 1000000

== Equal!= Not equal>= Greater than or equal tosize > 1000000

ch == ‘X’q

> Greater than

Important: The test for equality is ==, not =. This is the most common error in a C programa C program.


Example if statements < Less than<= Less than or equal to

E lif(age < 18)

printf(“Too young to vote!\n”);

== Equal!= Not equal>= Greater than or equal to> Greater than

if(area == 0)printf(“The plot is empty\n”);

else

> Greater than

elseprintf(“The plot has an area of %.1f\n”, area);

if(val < 0)( )printf(“Negative input is not allowed\n”);

else if(val == 0)printf(“A value of zero is not allowed\n”);

elseprintf(“The reciprocal is %.2f\n”, 1.0 / val);

Note the indentation


Note the indentation

Blocks Single Statementprintf(“This is a statement\n”);

{Block

{printf(“All items in a curly brace\n”);printf(“as if there are one statement”);printf(“They are executed sequentially”);

}}


Where is this useful? If the expression is true, all of the statements in

if(value > 0){

result = 1.0 / value;i tf("R lt %f\ " lt)

the block are executed

printf("Result = %f\n", result);}


Where is this useful? Will these two sections of code work

if(value > 0){

result = 1.0 / value;i tf("R lt %f\ " lt)

differently?

printf("Result = %f\n", result);}

if(value > 0)result = 1.0 / value;printf("Result = %f\n", result);printf( Result %f\n , result);


Where is this useful? Yes!

if(value > 0){

result = 1.0 / value;i tf("R lt %f\ " lt)printf("Result = %f\n", result);

}

if(value > 0)result = 1.0 / value;printf("Result = %f\n", result);

Will always execute!

printf( Result %f\n , result);


Nested Blocks What does this do?

if(bobsAge != suesAge) /* != means "not equal" */{

printf("Bob and Sue are different ages\n");if(b b A > A )if(bobsAge > suesAge){

printf("In fact, Bob is older than Sue\n");if((bobsAge ‐ 20) > suesAge)(( g ) g ){

printf("Wow, Bob is more than 20 years older\n");}

}}}


Importance of indentation See how much harder this is to read?

if(bobsAge != suesAge) /* != means "not equal" */{printf("Bob and Sue are different ages\n");if(b b A > A )if(bobsAge > suesAge){printf("In fact, Bob is older than Sue\n");if((bobsAge ‐ 20) > suesAge)(( g ) g ){printf("Wow, Bob is more than 20 years older\n");}}}}


Boolean Expressions

• An expression whose value is true or false• In C:

– integer value of 0 is “false”– nonzero integer value is “true”

• Example of Boolean expressions:

Relational operator– age < 40– graduation_year == 2010

Relational operator


#include <stdio.h>#include <stdbool.h>

Library that defines: bool, true, false

int main(){

const bool trueVar = true, falseVar = false; const int int3 3 int8 8;const int int3 = 3, int8 = 8;

printf("No 'boolean' output type\n");printf("bool trueVar: %d\n",trueVar);printf("bool falseVar: %d\n\n",falseVar);printf("int int3: %d\n",int3);printf("int int8: %d\n",int8);

}

What does the output look like?

}


#include <stdio.h>#include <stdbool.h>

Library that defines: bool, true, false

int main(){

const bool trueVar = true, falseVar = false; const int int3 3 int8 8;const int int3 = 3, int8 = 8;

printf("No 'boolean' output type\n");printf("bool trueVar: %d\n",trueVar);printf("bool falseVar: %d\n\n",falseVar);printf("int int3: %d\n",int3);printf("int int8: %d\n",int8);

}


}


// E l 3 ( ti d )// Example3 (continued…)

printf("\nint3 comparators\n");printf( \nint3 comparators\n );

printf("int3 == int8: %d\n",(int3 == int8));

printf("int3 != int8: %d\n",(int3!=int8));printf( int3 ! int8: %d\n ,(int3! int8));

printf("int3 < 3: %d\n",(int3 < 3));

printf("int3 <= 3: %d\n",(int3 <= 3));

Comparing values of two integer constants

printf("int3 > 3: %d\n",(int3 > 3));

printf("int3 >= 3: %d\n",(int3 >= 3));



// E l 3 ( ti d )// Example3 (continued…)

printf("\nint3 comparators\n");printf( \nint3 comparators\n );

printf("int3 == int8: %d\n",(int3 == int8));

printf("int3 != int8: %d\n",(int3!=int8));printf( int3 ! int8: %d\n ,(int3! int8));

printf("int3 < 3: %d\n",(int3 < 3));

printf("int3 <= 3: %d\n",(int3 <= 3));

printf("int3 > 3: %d\n",(int3 > 3));

printf("int3 >= 3: %d\n",(int3 >= 3));


More Examples

• char myChar = ‘A’; – The value of myChar==‘Q’ is false (0)

• Be careful when using floating point equality comparisons, especially with zero, e.g. myFloat==0


Suppose?

What if I want to know if a value is in a range?

Test for: 100 ≤ L ≤ 1000?


You can’t do… This code is WRONG and will fail.

if(100 <= L <= 1000){

printf(“Value is in range \n”);printf( Value is in range…\n );}


Why this fails… C Treats this code this way

if((100 <= L) <= 1000){

printf(“Value is in range \n”);printf( Value is in range…\n );}

Suppose L is 5000. Then 100 <= L is true, so (100 <= L) evaluates to true, which, in C, is a 1. Then it tests 1 <= 1000, which also returns true, even though you expected a false., g y p


Compound Expressions

• Want to check whether ‐3 <= B <= ‐1– Since B = ‐2, answer should be True (1)

• But in C, the expression is evaluated as– ((‐3 <= B) <= ‐1) (<= is left associative)– (‐3 <= B) is true (1)– (1 <= ‐1) is false (0)Th f i 0!– Therefore, answer is 0!


Compound Expressions

• Solution (not in C): (‐3<=B) and (B<=‐1)

• In C: (‐3<=B) && (B<=‐1)• Logical Operatorsg p

– And: &&– Or: ||– Not: !


Compound Expressions#i l d tdi h#include <stdio.h>

int main() {{const int A=2, B = ‐2;

printf("Value of A is %d\n", A);printf("0 <= A <= 5?: Answer=%d\n", (0<=A) && (A<=5));

printf("Value of B is %d\n", B);printf("‐3 <= B <= ‐1?: Answer=%d\n", (‐3<=B) && (B<=‐1));

}


Compound Expressions#include <stdio h>#include <stdio.h>

int main() {const int A=2, B = ‐2;

printf("Value of A is %d\n", A);printf("0 <= A <= 5?: Answer=%d\n" (0<=A) && (A<=5));printf( 0 <= A <= 5?: Answer=%d\n , (0<=A) && (A<=5));


Correct Answer!!!

} True (1) True (1)


Compound Expressions#include <stdio h>#include <stdio.h>

int main() {const int A=2, B = ‐2;

printf("Value of A is %d\n", A);printf("0 <= A <= 5?: Answer=%d\n" (0<=A) && (A<=5));printf( 0 <= A <= 5?: Answer=%d\n , (0<=A) && (A<=5));


Correct Answer!!!

} True (1) True (1)

&&

True (1)

&&


Truth TablesN A d O

p q !p p && q p || qNot And Or

True True

True FalseTrue False

False True

False False


Truth TablesN A d O

p q !p p && q p || q Not And Or

True True False

True False FalseTrue False False

False True True

False False True


Truth TablesN A d O


True True True

True False FalseTrue False False

False True False

False False False


Truth TablesN A d O


True True True

True False TrueTrue False True

False True True

False False False


Truth TablesN A d O

Our comparison operators: < <= == != >= >


True True False True True

True False False False TrueTrue False False False True

False True True False True

False False True False False


Precedence & AssociativityCan you guess what’s the answer?Can you guess what s the answer?

Relational operators have precedence and associativity(just like arithmetic operators)Use ( ) when in doubt


( ((A + B) > 5) && ( ((A=0) < 1) > ((A + B) – 2)) )

( (6 > 5) && ( ((A=0) < 1) > ((A + B) – 2)) )

( 1 && ( ( 0 < 1) > ((A + B) – 2)) )( 1 && ( ( 0 < 1) > ((A B) 2)) )

( 1 && ( 1 > ( 2 – 2) ) )

( 1 && ( 1 > 0 ) )

( 1 && 1 )( )

Answer: 1 Precedence: +/‐> <


&&

Associativity

“=“ is right associativeExample: X=Y=5

right associative: X = (Y=5)right associative: X (Y 5)expression Y=5 returns

value 5: X = 5


You should refer to theYou should refer to the C operator precedence and associative table

See for example, http://www.difranco.net/cop2220/op‐prec.htmp

Or just useOr just use parentheses whenever you’re unsure about

d dprecedence and associativity


Switch StatementA less general substitute for the multibranch if. It is used for selecting among discrete values (int), i.e. not continuous valuescontinuous values.

switch (int_expression){

case_list: statement list; _ ;

case_list: statement_list;

default:default: statement_list;

}


Switch & Functions

Behavior

• The int_expression is evaluated. If the value is in a case_list, execution begins at that statement_list and continues through subsequent statement_lists until: break, return or end of switchreturn, or end of switch.


Switch & Functions

#include <stdio.h>

void main() { int gender; printf("Enter your gender (male=1, female=2): ");

f("%d" & d )scanf("%d",&gender);

switch(gender) {{

case 1: printf("You are a male\n"); break;break;

case 2:printf("you are a female\n"); break;

default:printf("Not a valid input\n"); break;


} }

switch(gender) {

case 1: printf("You are a male\n"); break;

case 2:i f(" f l \ ")printf("you are a female\n");

break; default:printf("Not a valid input\n");printf( Not a valid input\n ); break;

}


Loops and Repetition

condition

true

false

statement

true

int count = 1

t 5true falsecount <= 5

statement

true

count ++


First, the tax programdouble income;int filingStatus;int numDependents;int children;int children;double standardDeduction;double deductions;double taxableIncome;double tax;

printf("Enter your annual income: ");scanf("%lf" &income);scanf( %lf , &income);

if(income < 9350){

printf("You may be poor, but you owe no taxes\n");exit(0);

}


printf("What is your filing status? \n1) single\n”);printf(“2) married filing jointly\n3) married filing separately\n");printf("Please enter a number: ");scanf("%d", &filingStatus);

switch(filingStatus){

case 1:numDependents = 1;standardDeduction = 5700;break;

case 2:printf("How many children do you have? ");printf( How many children do you have? );scanf("%d", &children);numDependents = children + 2;standardDeduction = 11400;break;

case 3:numDependents = 1;standardDeduction = 5700;break;break;

default:printf("Invalid input!\n");exit(1);b k


break;}

deductions = standardDeduction + numDependents * 3650;taxableIncome = income ‐ deductions;

if(taxableIncome < 0)( ){

tax = 0;}else if(taxableIncome <= 16750)( ){

tax = taxableIncome * 0.10;}else if(taxableIncome <= 68000)( ){

tax = 1675 + 0.15 * (taxableIncome ‐ 16750);}else if(taxableIncome <= 137300)( ){

tax = 9362.50 + 0.25 * (taxableIncome ‐ 68000);}else{

tax = 26687.50 + 0.28 * (taxableIncome ‐ 137300);}


printf("%.2f\n", tax);

Loops and Repetition

Loops in programs allow us to repeat blocks of code.p

Useful for:Trying again for correct inputCountingRepetitive activitiesRepetitive activitiesPrograms that never end


Three Types of Loops/Repetition in C • while

– top‐tested loop (pretest)

• for– counting loop– forever‐sentinel

• do– bottom‐tested loop (posttest)


The while loopd l ( )Top‐tested loop (pretest)

while (condition)statement;

Note that, as in IF selection, only one statement is executed. You need a block to repeat more than one statement (using { })


while(condition)statement;

conditionfalse

true

statement


Similar to the if statement

• Check the boolean condition• If true, execute the statement/block

Repeat the above until the boolean is falsep


Examplebool valid = true; // Until we know otherwise

printf("Enter the inductance in millihenrys: ");scanf("%lf", &l);

/* Test to see if the user entered an invalid value */ if(l <= 0){printf("You moron, you entered an invalid inductance!\n");valid = false;

}elseelseprintf("Okay, I guess that's reasonable\n");

Remember this? What if weRemember this? What if we input invalid values?


Example with while loop

bool valid = false; /* Until we know otherwise */

while(!valid) /* Loop until value is valid */{{


/* Test to see if the user entered an invalid value */ / /if(l < 0){

printf("You moron, you entered a negative inductance!\n");}}else if(l == 0){

printf("You are really dumb, you entered zero.\n");}}else{

printf("Okay, I guess that's reasonable\n");valid = true;;

}} What does this do different?


while

condition falsewhile (condition)

statement;

t t t

truewhile (condition){

statement1;statementstatement2;

}

hil (! lid) /* L til l i lid */while(!valid) /* Loop until value is valid */{


int i = 10;while(i > 0){

if(l > 0){

valid = true;}

}

{printf("i=%d\n", i);i = i ‐ 1;

}


}

1

Forever loops and never loops

Because the conditional can be “always true” or “always false”, you can get a loop that runs forever

t llor never runs at all.

int count=0;while(count !=0)

What is wrong with these statements?while(count !=0)

printf(“Hi Mom”);

while (count=1) //insidious error!!!while (count 1) //insidious error!!!count = 0;


How to count using while

First, outside the loop, initialize the counter variableTest for the counter’s value in the booleanDo the body of the loopLast thing in the body should change the value of the g y gcounter! i = 1;

while(i <= 10){{

printf("i=%d\n", i);i = i + 1;

}}


The for loop

The while loop is pretty general. Anything that can be done using repetition can be done with a while loop

Because counting is so common, there is a specialized construct called a for loop.

A for loop makes it easy to set up a counting loop


Three parts for(count=1;count<=5; count++)statement;

Three parts to a for loop (just like the while):• Set the initial value for the counterSet the initial value for the counter• Set the condition for the counter• Set how the counter changes each time through the• Set how the counter changes each time through the loop


for(count=1; count<=5; count++)printf(“count=%d\n”, count);printf( count %d\n , count);

count = 1

falsecount <= 5true

false

printf

count ++


Ascending for<=,++

for (control_var=init_value; control_var <=limit_value; control var++) statement;,

control_var = init_value

control_var++) statement;

true falsecontrol_var<= limit_value

true false

statement

control_var ++


Descending for>

for (control_var=init_value; control_var >=limit_value; control var ) statement;>=,‐‐

control_var = init_value

control_var‐‐) statement;

control_var>= limit_value

true false

statement

control_var ‐‐


Commentsd l l l• It is dangerous to alter control_var or limit_var

within the body of the loop.

• The components of the for statement can be a arbitrary statements e g the loop condition mayarbitrary statements, e.g. the loop condition may be a function call.


for(count=1; count<=5; count++)printf(“count=%d\n”, count); count = 1printf( count %d\n , count);

count <= 5truefalse

for(i=1; i<=10; i++){ count <= 5

printf

trueprintf("%d\n", i);}

for(t = 1.7; t < 3.5; t = t + 0.1)

count ++

( ; ; ){

printf("%f\n", t);}

for(i=1; i<5; i++){for(j=1; j<4; j++){

printf("%d * %d = %d\n", i, j, i * j);}

}


}

Top‐tested Equivalencep q

The following loopfor(x=init; x<=limit; x++)for(x=init; x<=limit; x++)

statement_listis equivalent tois equivalent to

x=init;while (x<=limit){while (x<=limit){

statement_list;x++;x++;

}


Some Magic Statements

s += 12; /* Equivalent to s = s + 12; */s ‐= 13; /* Equivalent to s = s – 13; */

These work fine for integers or floating point


break;

The break statement exits the containing loop immediately!while(true) /* Loop until value is valid */{{


/* Test to see if the user entered an invalid value *// Test to see if the user entered an invalid value / if(l <= 0){

printf("You moron, you entered an invalid inductance!\n");}}else{

printf("Okay, I guess that's reasonable\n");break;break;

}}


The do/while loop Often just called a “do loop”.

• do/while– bottom‐tested loop (posttest)

do{

angle += 2 * M PI / 20;angle + 2 M_PI / 20;sinVal = sin(angle);printf(“sin(%f) = %f\n”, angle, sinVal);

} while(sinVal < 0.5);} ( )


Bottom‐tested Loop: do

• Bottom‐tested (posttest)• One trip through loop is guaranteed, i.e.

i d lstatement is executed at least once

do do {statement

while (loop_condition);

{statement1;statement2;

}}while (loop_condition);

Usually!


do { statement; } while(condition)

statement;

conditiontrue false


do/while Examples i = 0;do

angle = M_PI / 2;do{

do{

i++;printf("%d\n", i);

} hil (i 10){angle ‐= 0.01;cosVal = cos(angle);printf("cos(%f)=%f\n", angle, cosVal);

} while(i < 10);

p ( ( ) \ , g , );} while(cosVal < 0.5);

do{

printf("Enter a value > 0: ");printf( Enter a value > 0: );scanf("%lf", &val);

} while(val <= 0);


Bottom‐tested Equivalence• Bottom tested do loop (posttest)• Bottom‐tested do loop (posttest)

do {

statement;}while (condition);( );

• Similar to bottom‐tested forever loopfor (;;){{

statement_list;if (!condition) break;

}


The “one off” error for(i=1; i<10; i++){

It is easy to get a for loop to be “one off” of the

{}

for(i 1; i< 10; i++)number you want. Be careful of the combination of init value

for(i=1; i<=10; i++){}

combination of init_valueand < vs. <= for(i=0; i<10; i++)

{

Counting from 0, with <, is a good combination

{}

is a good combination and good for invariants as well.


The “one off” error for(i=1; i<10; i++){ 9 l 1 9It is easy to get a for loop

to be “one off” of the

{}

for(i 1; i< 10; i++)

9 values: 1 to 9

number you want. Be careful of the combination of init value

for(i=1; i<=10; i++){}

10 values: 1 to 10combination of init_valueand < vs. <= for(i=0; i<10; i++)

{

Counting from 0, with <, is a good combination

{} 10 values: 0 to 9

is a good combination and good for invariants as well.


while, for, do/whilefor(i=1; i<=10; i++){

printf("%d\n", i);}

for(t 1 7; t < 3 5; t t + 0 1)for(t = 1.7; t < 3.5; t = t + 0.1){

printf("%f\n", t);}

do{

printf("Enter a value > 0: ");scanf("%lf" &val);

i = 0;do{

i++;printf("%d\n" i);

scanf( %lf , &val);} while(val <= 0);

for(i=1; i<5; i++){

hil (! lid) /* L til l i lid */

printf( %d\n , i);} while(i < 10); for(j=1; j<4; j++)

{printf("%d * %d = %d\n", i, j, i * j);

}} i t i 10while(!valid) /* Loop until value is valid */

{printf("Enter the inductance in millihenrys: ");scanf("%lf", &l);

if(l > 0) angle = M_PI / 2;

} int i = 10;while(i > 0){

printf("i=%d\n", i);i = i ‐ 1;if(l > 0)

{valid = true;

}}

do{angle ‐= 0.01;cosVal = cos(angle);printf("cos(%f)=%f\n" angle cosVal);

;}


printf( cos(%f)=%f\n , angle, cosVal);} while(cosVal < 0.5);

3

Functions


Moon Landings

Time Fuel Velocity

Will Cyr 13 86.20 ‐0.49

Yongjiao Yu 13 86.00 ‐1.82

Bin Tian 13 87 00 ‐1 69Bin Tian 13 87.00 ‐1.69

Nan Xia 13 87.00 ‐1.69

Chenli Yuan 13 87.00 ‐1.69

Scott Oliver 13 87.70 ‐2.74

Mike Robell 13 87.88 ‐3.00


Triangle Area Computation 12 ppa −=

p2=(x2 y2)

p3=(x3,y3)23

13

ppc

ppb

−=

−=c

p2=(x2,y2)

ab

22 )12()12( yyxxa −+−=

p1=(x1,y1) ))()((cba

cpbpapparea++

−−−=

2cbap ++

=

How would you write this program?


Variablesint main(){

double x1=0 y1=0;double x1=0, y1=0;double x2=17, y2=10.3;double x3=‐5.2, y3=5.1;

double a, b, c; /* Triangle side lengths */double p; /* For Heron's formula */double area;

12 ppa −=

23

13

ppc

ppb

pp

−=

−=

pp

22 )12()12( yyxxa −+−=


Lengths of Edges

a = sqrt((x1 ‐ x2) * (x1 ‐ x2) + (y1 ‐ y2) * (y1 ‐ y2));b = sqrt((x1 ‐ x3) * (x1 ‐ x3) + (y1 ‐ y3) * (y1 ‐ y3));c = sqrt((x2 ‐ x3) * (x2 ‐ x3) + (y2 ‐ y3) * (y2 ‐ y3));

13

12

ppb

ppa

−=

−=

23 ppc −=

22p2=(x2,y2)

p3=(x3,y3)c

22 )12()12( yyxxa −+−=

ab

))()(( cpbpapparea =p1=(x1,y1)

2

))()((cbap

cpbpapparea++

=

−−−=


Areap = (a + b + c) / 2;area = sqrt(p * (p ‐ a) * (p ‐ b) * (p ‐ c));q (p (p ) (p ) (p ));

printf("%f\n", area); ))()(( cpbpapparea −−−= ))()(( cpbpapparea −−−=

2cbap ++

=2

cbap ++=


Whole Programint main(){

double x1=0, y1=0;Wh if I d i kdouble x2=17, y2=10.3;

double x3=‐5.2, y3=5.1;

double a, b, c; /* Triangle side lengths */

What if I made a mistake on the edge length equation?

double a, b, c; / Triangle side lengths /double p; /* For Heron's formula */double area;

t(( 1 2) * ( 1 2) ( 1 2) * ( 1 2))a = sqrt((x1 ‐ x2) * (x1 ‐ x2) + (y1 ‐ y2) * (y1 ‐ y2));b = sqrt((x1 ‐ x3) * (x1 ‐ x3) + (y1 ‐ y3) * (y1 ‐ y3));c = sqrt((x2 ‐ x3) * (x2 ‐ x3) + (y2 ‐ y3) * (y2 ‐ y3));

p = (a + b + c) / 2;area = sqrt(p * (p ‐ a) * (p ‐ b) * (p ‐ c));

printf("%f\n" area);printf( %f\n , area);}


Functions

• Functions are subprograms that perform some operation and return one value

• They “encapsulate” some particular operation, so it can be re‐used by others (for example, the abs() or

() f i )sqrt() function)


Characteristics

• Reusable code– code in sqrt() is reused often

• Encapsulated code– implementation of sqrt() is hidden

• Can be stored in libraries– sqrt() is a built‐in function found in the math library


Writing Your Own Functions

• Consider a function that converts temperatures in Celsius to temperatures in Fahrenheit.– Mathematical Formula:

F C * 1 8 32 0F = C * 1.8 + 32.0

– We want to write a C function called CtoF– We want to write a C function called CtoF


Convert Function in C

double CtoF ( double paramCel ) {

return paramCel*1.8 + 32.0; }

• This function takes an input parameter called paramCel (temp in degree Celsius) and returns aparamCel (temp in degree Celsius) and returns a value that corresponds to the temp in degree Fahrenheita e e


How to use a function?

#include <stdio.h>

double CtoF( double );

/************************************************************************* Purpose: to convert temperature from Celsius to Fahrenheit ************************************************************************/int main() {

double c, f;printf(“Enter the degree (in Celsius): “);scanf(“%lf”, &c);

f = CtoF(c);printf(“Temperature (in Fahrenheit) is %lf\n”, f);

}

double CtoF ( double paramCel) {

return paramCel * 1.8 + 32.0;}}


TerminologyD l ti d bl Ct F( d bl )• Declaration: double CtoF( double );

• Invocation (Call): Fahr CtoF(Cel);• Invocation (Call): Fahr = CtoF(Cel);

• Definition:• Definition: double CtoF( double paramCel ) {{ return paramCel*1.8 + 32.0;

}}


Function Declarationl ll d f• Also called function prototype:

return_type function_name (parameter_list)

double CtoF(double)• Declarations describe the function:

– the return type and function name

( )

– the type and number of parameters


Function Definition

return_type function_name (parameter_list) {

….function body….

}

double CtoF(double paramCel) {{

return paramCel*1.8 + 32.0;}


Function Invocation1 C ll i

int main() { …f = CtoF(c);

1. Call copies argument c to

tf CtoF(c); } parameter

paramCel

double CtoF ( double paramCel ){

2. Control f {

return paramCel*1.8 + 32.0;}

transfers to function “C F”“CtoF”


Invocation (cont)

int main() { …

f = CtoF(c); }

3. Expression in “CtoF” is

} evaluated

double CtoF ( double paramCel ){

return paramCel*1 8 + 32 0;

4. Value of expression return paramCel*1.8 + 32.0;

}expression is returned to “main”to main


Local Objects

• The parameter “paramCel” is a local object which is defined only while the function is executing. Any attempt to use “paramCel” outside the function is an error.

• The name of the parameter need not be the same as th f th t T tthe name of the argument. Types must agree.


Can we do better than this?

int main(){double x1=0, y1=0;double x2 17 y2 10 3double x2=17, y2=10.3;double x3=‐5.2, y3=5.1;

double a, b, c; /* Triangle side lengths */double p; /* For Heron's formula */double area;

a = sqrt((x1 x2) * (x1 x2) + (y1 y2) * (y1 y2));a = sqrt((x1 ‐ x2) (x1 ‐ x2) + (y1 ‐ y2) (y1 ‐ y2));b = sqrt((x1 ‐ x3) * (x1 ‐ x3) + (y1 ‐ y3) * (y1 ‐ y3));c = sqrt((x2 ‐ x3) * (x2 ‐ x3) + (y2 ‐ y3) * (y2 ‐ y3));


printf("%f\n" area);printf( %f\n , area);}


What should we name our function?a = sqrt((x1 ‐ x2) * (x1 ‐ x2) + (y1 ‐ y2) * (y1 ‐ y2));

“Length” sounds like a good ideaLength sounds like a good idea.

??? Length( ??? ){{}


What does our function need to know?a = sqrt((x1 ‐ x2) * (x1 ‐ x2) + (y1 ‐ y2) * (y1 ‐ y2));

(x, y) for two different points:

??? Length(double x1, double y1, double x2, double y2)

{{}


What does our function return?a = sqrt((x1 ‐ x2) * (x1 ‐ x2) + (y1 ‐ y2) * (y1 ‐ y2));

A computed value which is of type double

double Length(double x1, double y1, double x2, double y2)

{{}


How does it compute it?a = sqrt((x1 ‐ x2) * (x1 ‐ x2) + (y1 ‐ y2) * (y1 ‐ y2));

A computed value which is of type double

double Length(double x1, double y1, double x2, double y2)

{{double len;len = sqrt((x1 ‐ x2) * (x1 ‐ x2) + (y1 ‐ y2) * (y1 ‐ y2));return(len);return(len);

}


Using This#include <stdio.h>#include <math.h>

/* Declaration */double Length(double x1, double y1, double x2, double y2);

/* * Program to determine the area of a triangle*/

int main(){

Declaration

{double x1=0, y1=0;double x2=17, y2=10.3;double x3=‐5.2, y3=5.1;

double a, b, c; /* Triangle side lengths */d bl /* ' f l */double p; /* For Heron's formula */double area;

a = Length(x1, y1, x2, y2);b = Length(x1, y1, x3, y3);c = Length(x2, y2, x3, y3);

Invocations


printf("%f\n", area);}}

/* Definition */double Length(double x1, double y1, double x2, double y2){

double len;len = sqrt((x1 ‐ x2) * (x1 ‐ x2) + (y1 ‐ y2) * (y1 ‐ y2));

Definition


len sqrt((x1 x2) (x1 x2) (y1 y2) (y1 y2));return(len);

}

1

Potential Errors#include <stdio.h>double convert( double );

int main()int main() {

double c, f;printf(“Enter the degree (in Celsius): “);scanf(“%lf”, &c);scanf( %lf , &c); f= convert(c);printf(“Temp (in Fahrenheit) for %lf Celsius is %lf”, paramCel, f);

} Error! paramCel is not defined

double CtoF( double paramCel) {

return c * 1.8 + 32.0;} Error! C is not defined

Scope – Where a variable is known to existError! C is not defined known to exist.

No variable is known outside of the curly braces that contain it, even if the same name is used!


Another Example

#include <stdio.h>

double GetTemperature();p ();double CelsiusToFahrenheit( double );void DisplayResult( double, double );

int main()

Declarationsint main(){

doubleTempC, // Temperature in degrees CelsiusTempF; // Temperature in degrees Fahrenheit

TempC = GetTemperature();TempF = CelsiusToFahrenheit(TempC); InvocationsTempF = CelsiusToFahrenheit(TempC);DisplayResult(TempC, TempF);

return 0;}


Function:GetTemperature

double GetTemperature(){

double Temp;p;

printf("\nPlease enter a temperature in degrees Celsius: ");scanf("%lf", &Temp);return Temp;return Temp;

}


Function: CelsiusToFahrenheitdouble CelsiusToFahrenheit(double Temp){

return (Temp * 1.8 + 32.0);}}


Function: DisplayResultvoid DisplayResult(double CTemp, double FTemp){

printf("Original: %5.2f C\n", CTemp);printf("Equivalent: %5 2f F\n" FTemp);printf( Equivalent: %5.2f F\n , FTemp);

return;}


Declarations (Prototypes)

double GetTemp( ); double CelsiusToFahrenheit( double );double CelsiusToFahrenheit( double );void Display( double, double );

• voidmeans “nothing”. If a function doesn’t return a gvalue, its return type is void


Abstraction 1. Get Temperature2. Convert Temperature

int main(){doubleTempC // Temperature in degrees Celsius

p3. Display Temperature

TempC, // Temperature in degrees CelsiusTempF; // Temperature in degrees Fahrenheit

TempC = GetTemperature();( )TempF = CelsiusToFahrenheit(TempC);

DisplayResult(TempC, TempF);

return 0; We are hiding details on how somethingreturn 0;}

We are hiding details on how something is done in the function implementation.


Another Way to Compute Factorial

Pseudocode for factorial(n)

if n == 0 thenresult = 1

else result = n * factorial(n – 1)result = n factorial(n – 1)

After all, 5! = 5 * 4 * 3 * 2 * 1 = 5 * 4!


Factorial function contains an invocation of itself.

Recursive Functionsinvocation of itself.

We call this a: recursive call.

int Factorial(int n){

if(n == 0)t 1return 1;

else return n * Factorial(n‐1);

}

Recursive functions must have a base case

}

This works much like proof by induction (if n == 0): why?proof by induction.

if n == 0 thenif n 0 thenresult = 1

else result = n * factorial(n 1)


result = n * factorial(n – 1)

Infinite Recursion What if I omit the “base case”?


return n * Factorial(n‐1);}

This leads to infinite recursion!

}

Factorial(3)=

cbowen@ubuntu:~/cse251$ ./combi1Input n: 5Input k: 3Factorial(3)=

3 * Factorial(2) = 3 * 2 * Factorial(1) = 3 * 2 * 1 * Factorial(0) =

Input k: 3Segmentation fault

3 2 1 Factorial(0) 3 * 2 * 1 * 0 * Factorial(‐1) = …


Psuedocode and Function if n == 0 thenresult = 1


if(n == 0)t 1

else result = n * factorial(n – 1)Base Case

return 1;else

return n * Factorial(n‐1);}}

Declaration: int Factorial(int n);

Invocation: f = Factorial(7);Invocation: f Factorial(7);

Definition:


if(n == 0)freturn 1;

else return n * Factorial(n‐1);

}


Pointers and Reference parameters


Today’s Lecture

Pointers are variables that Address Variable Value

0 4520 B 181 store memory addresses0x4520 B 181

0x4524 pB 0x4520

Address Variable ValueBefore we learn about

Memory Address

0x4520 B 181

0x4524 pB 0x4520

pointers, we must learn more about addresses

Pointer Value


Computer Memory

Memory is just a long list of numbers one after the otherMy laptop has over 4 Billion of these numbersthese numbers

Each number is 8 bits (BYTE)We combine them to make

integers and floating point valuesintegers and floating point values


Computer Memory

int (21) float (18.2)


Memory AddressesMemory addresses in computers are often 32 bits (or nowadays, 64‐bits) long, e.g.

01111111111111111111101010001100

Another way to represent an address is to use hexadecimal:

0x 7ffffa8c


Hexadecimal (Base‐16) 0000 = 0 1000 = 80001 = 1 1001 = 90010 = 2 1010 = 10 = a0011 = 3 1011 = 11 = b

I have included this chart on your

0100 = 4 1100 = 12 = c0101 = 5 1101 = 13 = d0110 6 1110 14

yworksheet so you can refer to it.

0110 = 6 1110 = 14 = e0111 = 7 1111 = 15 = f


Addressesb dd ( )32‐bit address (Binary):

0111 1111 1111 1111 1111 1010 1000 11007 f f f f a 8 c

32‐bit address (Hex): 0x 7 f f f f a 8 cNotes:

– In C “0x” indicates a Hexadecimal number– Convert every four bits to a hex digit


Arithmetic (in Hex) 0000 = 0 1000 = 80001 = 1 1001 = 9

Sum: 0x 90 + 0x 04

0001 1 1001 90010 = 2 1010 = 10 = a0011 = 3 1011 = 11 = b

0x 94 0100 = 4 1100 = 12 = c0101 = 5 1101 = 13 = d0110 = 6 1110 = 14 = e0111 = 7 1111 = 15 = f


Arithmetic (in Hex) 0000 = 0 1000 = 80001 = 1 1001 = 9

Sum with carry: 0x 8c + 0x 04

0001 1 1001 90010 = 2 1010 = 10 = a0011 = 3 1011 = 11 = b

0x ?? 0100 = 4 1100 = 12 = c0101 = 5 1101 = 13 = d0110 = 6 1110 = 14 = e0111 = 7 1111 = 15 = f


Arithmetic (in Hex)

Sum with carry: 0x 8c + 0x 04 0x ??

– What is “c + 4”? – In decimal it is “12 + 4 = 16”

hi h i H “10” (0 d 1)which is Hex “10” (0 and carry 1)


Arithmetic (in Hex)1

Sum with carry: 0x 8c + 0x 04 0x ??




Arithmetic (in Hex)1

Sum with carry: 0x 8c + 0x 04 0x 90




Bytes and Words

Remember

byteword

8 bits = 1 byte32 bits = 4 bytes = 1 word.

32‐bit address machines are addressed by bytes so

consecutive words haveconsecutive words have addresses that differ by four


32‐bit AddressesH h i 32 bi dd (i H ) f dHere are three consecutive 32‐bit addresses (in Hex) of words:

0x00712050 dc bb 21 000x00712054 01 00 00 000x00712058 00 00 00 00


PointersPointers are variables that contain addresses

Just like other variables, they must be declared before being usedbeing used

Declaration: int *p; /* instead of int p for integers */

int * means p is a pointer variable that stores the addressint * means p is a pointer variable that stores the address of an integer variable


Pointers and Ref parameters

Pointer Initialization

Declaration: int a = 2; /* a is an integer */i t * A & /* A i i t t i iint *pA = &a;/* pA is a pointer containing

the address of a */

“&” operator means “address of”Read it as “at”Read it as “at”


The Address Game int a = 2; int *pA = &a;

a

pAp

An Intel processor is called a little endianprocessor because it stores values with the leastprocessor because it stores values with the least significant byte first. You read it in reverse order.

0x00602104 in memory will be: 04 21 60 00


Example Program“%x” prints the hexadecimal value

Operators:

int a = 21;int *pA = &a;

p& “address of”* “dereference”

printf("%d\n", a);printf("%x\n", a);printf("% \n" &a);

2115f

Output

printf("%x\n", &a);printf("%x\n", pA);printf("%d\n", *pA);printf("%x\n", &pA);

bfee861cbfee861c21bfee8618p ( p ) bfee8618


#include<stdio.h>

int main() {{

int a = 15, b = 38;int *c = &a;printf("%x : %d\n", &a, a);p ( \ , , );printf("%x : %d\n", &b ,b);printf("%x : %x : %d\n", &c, c, *c);

a = 49;a = 49;printf("%x : %d\n", &a, a);printf("%x : %d\n", &b ,b);printf("%x : %x : %d\n", &c, c, *c);p ( \ , , , );

c = &b;printf("%x : %d\n", &a, a);printf("%x : %d\n" &b b);printf( %x : %d\n , &b ,b);printf("%x : %x : %d\n", &c, c, *c);

}


#include<stdio.h>

int main() {{




}


First SectionDeclares a and b as integersDeclares c as a pointer that contains the address of a (“points to a”)

int a = 15 b = 38;Address Memory Name

address of a ( points to a )

int a = 15, b = 38;int *c = &a;printf("%x : %d\n", &a, a);

0xeffffa94

0 ffff 90

a

b

15

38printf("%x : %d\n", &b ,b);printf("%x : %x : %d\n", &c, c, *c);

0xeffffa90

0xeffffa8c

b

c

38

0xeffffa94



int a = 15 b = 38;Address Memory Name


int a = 15, b = 38;int *c = &a;printf("%x : %d\n", &a, a);

0xeffffa94

0 ffff 90

a

b

15

38printf("%x : %d\n", &b ,b);printf("%x : %x : %d\n", &c, c, *c);

0xeffffa90

0xeffffa8c

b

c

38

0xeffffa94

Output:effffa94 15effffa90 38effffa8c effffa94 15



int a = 15 b = 38;


Note the difference betweenint a = 15, b = 38;int *c = &a;printf("%x : %d\n", &a, a);

Note the difference between *C in variable declaration and *C in printf

printf("%x : %d\n", &b ,b);printf("%x : %x : %d\n", &c, c, *c);



#include<stdio.h>

int main() {{




}


Second SectionAddress Memory Namey

0xeffffa94 A

B

15 49

38

a = 49;printf("%x : %d\n", &a, a);printf("%x : %d\n", &b ,b);

0xeffffa90

0xeffffa8c

B

C

38

0xeffffa94

printf("%x : %x : %d\n", &c, c, *c);


Second ExampleAddress Memory Namey

0xeffffa94 A

B

15 49

38

a = 49;printf("%x : %d\n", &a, a);printf("%x : %d\n", &b ,b);

0xeffffa90

0xeffffa8c

B

C

38

0xeffffa94

printf("%x : %x : %d\n", &c, c, *c);



#include<stdio.h>

int main() {{




}


Third Section Address Memory Name

0xeffffa94

0xeffffa90

A

B

49

38

c = &b; /* c now points to b */printf("%x : %d\n", &a, a);printf("%x : %d\n", &b ,b);i f("% % %d\ " & * )

0xeffffa8c C0xeffffa90printf("%x : %x : %d\n", &c, c, *c);


Third Section Address Memory Name

0xeffffa94

0xeffffa90

A

B

49

38

c = &b; /* c now points to b */printf("%x : %d\n", &a, a);printf("%x : %d\n", &b ,b);i f("% % %d\ " & * )

0xeffffa8c C0xeffffa90printf("%x : %x : %d\n", &c, c, *c);

O t tOutput:effffa94 49effffa90 38effffa8c effffa90 38effffa8c effffa90 38


Reference parameters

A valuable use for pointers:Passing addresses to a function


Argument & Returned Value

Consider a function call y=f(x). – The value x is passed to the function f – A value is returned and assigned to y.

d h h l f– By passed we mean that the value of argument x is copied to the parameter in the function. Some calculation is performed and the result is returned and passigned to y.


Example

int x, y;x = 5;y = Square(x);

Address Memory Namey = Square(x);

int Square(int t) {

0xeffffa94 … x

y0xeffffa98{ return t*t

}…

… y0xeffffa98

}…

…


Example


Address Memory Namey = Square(x);

int Square(int t) {

0xeffffa94 5 x


}…

… y0xeffffa98

}…

…


ExampleThe call Square(x) :creates a variable tcopies the value of x to t


Address Memory Name

copies the value of x to t

y = Square(x);

int Square(int t) 0xeffffa94 5 x


}…

… y0xeffffa98

}…

50xeffffc8c t0xeffffc8c


ExampleThe call Square(x) :creates a variable tcopies the value of x to t

int x, y;x = 5;y = Square(x); Address Memory Name

pcalculates t * treturns t

y = Square(x);

int Square(int t)

y

0xeffffa94 5 x


}…

… y0xeffffa98

}…

50xeffffc8c t0xeffffc8c

0xeffffc90 25 temp


y=f(x)

Only one valued returned What if we want to return more than one value?

– Solution is to use pointers to variables in the calling function


How to do this in C

The approach is to pass the address (using the & operator) of the value to be modified.

We call such a parameter a reference parameter.

Use the * operator to change the reference parameter value


Function Reference Params

int val = 10;MyFun(&val);

Name Address Value

val 0xeffffa90 10MyFun(&val);printf(“%d”,val);

void MyFun(int *param){

*param = 27;}




Name Address Value



Name Address Value

param 0xefffea88 0xefffa90 *param = 27;}

param 0xefffea88 0xefffa90




Name Address Value



Name Address Value

param 0xefffea88 0xefffa90 *param = 27;}

param 0xefffea88 0xefffa90




Name Address Value



Prints: 27

The memory used by *param = 27;

}the function is destroyed when it returnsreturns.


What will this do different?

int val = 10;MyFun2(val);

Name Address Value

val 0xeffffa90 10MyFun2(val);printf(“%d”,val);

void MyFun2(int param){

param = 27;}


Cards program/* C t d d d it *//* Create a random card and suit *//* This will compute a random number from 0 to 3 */suit1 = rand() % 4; This program repeats code. We don’t

like to do that. But, we could not put h d d i f i b/* This will compute a random number from 1 to 13 */

card1 = rand() % 13 + 1;

do

the card draw into a function because a function can only return one value, or so we thought!

do{

/* Create a random card and suit *//* This will compute a random number from 0 to 3 */suit2 = rand() % 4;

/* This will compute a random number from 1 to 13 */card2 = rand() % 13 + 1;card2 = rand() % 13 + 1;

} while(card1 == card2 && suit1 == suit2);


Solution, pass by reference using pointers/* Create a random card and suit */DrawCard(&card1, &suit1);

do{

DrawCard(&card2, &suit2);} while(card1 == card2 && suit1 == suit2);

void DrawCard(int *card, int *suit){

/* This will compute a random number from 0 to 3 */

Don’t forget:

*suit← to set the value random number from 0 to 3 /*suit = rand() % 4;

/* This will compute a random number from 1 to 13 */

suit ← to set the value&card1 ← to get the address

Pass with & number from 1 to 13 /*card = rand() % 13 + 1;

}Set with *


Intro to Arrays

Storing List of Data


Why ArraysSuppose we want to store theSuppose we want to store the grade for each student in a class

/* Need a variable for each? *// Need a variable for each? /int bob, mary, tom, …;

W bWow, cumbersome…

Easier to have a variable that stores the grades for all students


An array is a “Chunk of memory”f hAn array is a contiguous piece of memory that can

contain multiple valuesTh l ithi th ti h k bThe values within the contiguous chunk can be addressed individually

Add i

grades 74 59 95 85 71 45 99 82 76

0xeffffa00 0xeffffa04 0xeffffa08 0xeffffa0c 0xeffffa10 0xeffffa14 0xeffffa18 0xeffffa1c 0xeffffa20Address in memory


Array: “Chunk of memory”

grades 74 59 95 85 71 45 99 82 76

0xeffffa00 0xeffffa04 0xeffffa08 0xeffffa0c 0xeffffa10 0xeffffa14 0xeffffa18 0xeffffa1c 0xeffffa20

Physical address

index 0 1 2 3 4 5 6 7 8

Use an index to access individual elements of the array:grades[0] is 74, grades[1] is 59, grades[2] is 95, and so on


Array Declaration

Syntax for declaring array variable: type array_name[capacity];

– type can be any type (int, float, char, …)– array_name is an identifier– capacity is the number of values it can store (indexing starts at 0)starts at 0)


ExampleNotice: The first location is location 0 (zero)!

int x[3]; // an array of 3 integers double y[7]; // an array of 7 doubles

Storage, e.g. 4‐bytes per int

X

0 1 2

First Location


Operations with Arrays

• Assignment: – x[0] = 6; /* Assign 6 to element x[0] */

[2] 3 1 /* A i 3 1 l [2] */– y[2] = 3.1; /* Assign 3.1 to element y[2] */

• AccessAccess– m = x[2];– p = y[0];

• Input/Output: the elements are handled as their types e g– the elements are handled as their types, e.g.

scanf(“%d %lf”, &x[2], &y[3]); printf(“%d %lf\n”,x[0], y[2]); /* output 6 and 3.1 */


Arithmetic Operations()int main()

{double x[5];

Variable Declarationfor the arraydouble x[5];

x[0] = 1;

for the array

[ ] ;x[1] = 2;x[2] = x[0] + x[1]; /* X[2] = 3 */x[3] = x[2] / 3; /* X[3] = 1 */x[4] = x[3] * x[2]; /* X[4] = 3 */

}}


for loops

“for” loops are ideal for processing elements in the array.

int main() {

int i;double values[4] = {3 14 1 0 2 61 5 3};double values[4] = {3.14, 1.0, 2.61, 5.3};double sumValues = 0.0;

for (i=0; i<4; i++) {

sumValues = sumValues + values[i];}printf(“Sum = %lf\n” sumValues);printf( Sum %lf\n , sumValues);

}


for loops

“for” loops are ideal for processing elements in the array.int main() {

int i;double values[4] = {3 14 1 0 2 61 5 3};double values[4] = {3.14, 1.0, 2.61, 5.3};double sumValues = 0.0;

for (i=0; i<=4; i++)

ERROR! Out of bound

{sumValues = sumValues + values[i];

}printf(“Sum = %lf\n” sumValues);printf( Sum %lf\n , sumValues);

}


Initialization

Syntax: int X[4] = {2, 4, 7, 9};Behavior: initialize elements starting with leftmost, ehavior: initiali e elements starting with leftmost,i.e. element 0. Remaining elements are initialized to zero. 2 4 7 9X 2 4 7 9X

0 1 2 3

Initialize all to 0: int X[4]={0};


Exampleint main(){double grades[5] = {90, 87, 65, 92, 100};g [ ] { , , , , };double sum;int i;

printf("The first grade is: % 1f\n" grades[0]);printf("The first grade is: %.1f\n", grades[0]);

sum = 0;for(i=0; i<5; i++){sum += grades[i];

}printf("The average grade is: % 1f\n" sum / 5);printf( The average grade is: %.1f\n , sum / 5);

grades[2] = 70; /* Replaces 65 */grades[3] = grades[4]; /* Replaces 92 with 100 */

}


Constants for capacityGood programming practice: use #define for constants in your program

For example: #define MaxLimit 25#define MaxLimit 25

int grades[MaxLimit]; f (i t i i M Li it i ){ }for(int i; i<MaxLimit; i++){ };

If size needs to be changed only the capacityIf size needs to be changed, only the capacity “MaxLimit” needs to be changed.


Arrays as parameters of functionsi t i ()int main() {

double values[4] = {3.14, 1.0, 2.61, 5.3};[ ] { , , , };

printf(“Sum = %lf\n”, SumValues( values, 4));}}

Suppose we want a function that sums up values of the array


Arrays as parameters of functionsdouble SumValues(double x[], int numElements){

int i;double result = 0;for (i=0; i < numElements; i++)for (i=0; i < numElements; i++)

result = result + x[i];return result;

“[]” flags the parameter as an array.– ALWAYS passed by reference

return result;}

ALWAYS passed by referenceArray size is passed separately (as numElements)


Example

Program Behavior1. Create an array of random numbers2. Print unsorted array3. Sort the array

d4. Print sorted array


Sample output

Array before sortingElement 0 : 58.7000Element 1 : 8.0100Element 2 : 72.3700 p p

The array elements are randomly

Element 3 : 4.6500Element 4 : 58.3000Element 5 : 92.1700Element 6 : 95.3100

generatedElement 7 : 4.3100Element 8 : 68.0200Element 9 : 72.5400

Array after sortingElement 0 : 4.3100Element 1 : 4.6500Element 2 : 8.0100Element 3 : 58.3000Element 4 : 58.7000Element 5 : 68.0200Element 6 : 72.3700Element 7 : 72.5400Element 8 : 92.1700Element 9 : 95.3100


#include <stdio.h>#include <stdlib.h> Functions are your friends!

void PrintArray( double [], int );void SortArray( double [] int );

Functions are your friends! Make them work and then use them to do work!

void SortArray( double [], int );void Swap (double *, double *);


#define NumElements 10

int main() {{

int i;double values[NumElements]; /* The array of real numbers */

srand(time(NULL));( ( ));

for (i=0; i < NumElements; i++){

values[i] = (double)(rand() % 10000) / 100.0;[ ] ( )( () ) / ;}

printf("\nArray before sorting\n");PrintArray( values, NumElements );y( , );

SortArray( values, NumElements );

printf("\nArray after sorting\n");p ( y g );PrintArray( values, NumElements );

return 0;}


}


int main() {

Array declarationDeclare an array of 10 doubles{


srand(time(NULL));

Declare an array of 10 doublesThe indices range from 0 to 9,

i.e. Value[0] to Value[9]( ( ));



printf("\nArray before sorting\n");PrintArray( values, NumElements );y( , );


printf("\nArray after sorting\n");p ( y g );PrintArray( values, NumElements );

return 0;}


}


int main() {{


srand(time(NULL));( ( ));



printf("\nArray before sorting\n");PrintArray( values, NumElements );

Initialize the array with random valuesrand() returns a pseudo random number between 0 andy( , );


printf("\nArray after sorting\n");

rand() returns a pseudo random number between 0 and RAND_MAXrand()%10000 yields a four-digit integer remainder/100 0 moves the decimal point left 2 placesp ( y g );

PrintArray( values, NumElements );

return 0;}

/100.0 moves the decimal point left 2 placesSo, Values is an array of randomly generated 2-decimal digit numbers between 0.00 and 99.99


}

printf("\nArray before sorting\n");PrintArray( values, NumElements );

PrintArray prints the elements of the array in the order they are given to it

SortArray( values, NumElements ); SortArray sorts the elements into ascending order

printf("\nArray after sorting\n");PrintArray( values, NumElements );

order


Parameter Passing

void PrintArray( double array[], int size ) {} array is a C array of doubles

array is passed by reference, i.e. any changes to parameter array in the function would change the argument valueschange the argument valuesThe array size is passed as “size”size


void PrintArray( double array[], int size ) {

i t iint i;

for (i=0; i<size; i++)printf(" Element %5d : %8 4lf\n" i array[i]);

array[i] is a double so the output needs to be “%f”

printf( Element %5d : %8.4lf\n ,i, array[i]);}

array[i] is a double so the output needs to be %f

The range of the “for” statement walks through theThe range of the for statement walks through the whole array from element 0 to element N‐1.


Sorting Arrayvoid SortArray( double array[], int size){}}

array is an array of doubles.

i d b f i harray is passed by reference, i.e. changes to parameter array change the argument valuesThere is no size restriction on array so the size is passed as “size”.


Selection Sort8 2 6 4array

0 1 2 3



0 1 2 3

Search from array[0] to array[3] to find the smallest number


Selection SortSearch from array[0] to array[3]

8 2 6 4array

0 1 2 3

y[ ] y[ ]to find the smallest number andswap it with array[0]

2 8 6 4

0 1 2 30 1 2 3



0 1 2 3

2 8 6 4

0 1 2 3

Search from array[1] to array[3] to find the smallest number

0 1 2 3



0 1 2 3

2 8 6 4

0 1 2 3

Search from array[1] to array[3] to find the smallest number andswap it with array[1]0 1 2 3 swap it with array[1]

2 4 6 8

0 1 2 3



0 1 2 3

2 8 6 4

0 1 2 30 1 2 3

Search from array[2] to array[3] to find the smallest number and2 4 6 8swap it with array[2]

0 1 2 3



0 1 2 3

2 8 6 4

0 1 2 30 1 2 3

Search from array[2] to array[3] to find the smallest number and 2 4 6 8swap it with array[2]

0 1 2 3

2 4 6 8 A d d !2 4 6 8

0 1 2 3

And we are done!


Selection Sort How many iterations are there?Answer: 3 ( from i = 0 to i = 2)

8 2 6 4array

0 1 2 3

Answer: 3 ( from i 0 to i 2)

More generally, if number of elements in the array is size

2 8 6 4

0 1 2 3

elements in the array is size,you need to iterate fromi = 0 to i = size ‐ 2

0 1 2 3

2 4 6 8

0 1 2 3

2 4 6 82 4 6 8

0 1 2 3


Selection Sort At every iteration i, you need to search f [i] t [ i 1] tfrom array[i] to array[size – 1] to find the smallest element

How to do this?

2 8 6 4

0 1 2 3How to do this?


Selection Sort At every iteration i, you need to search from array[i] to array[size – 1] to

2 8 6 4

0 1 2 3

find the smallest element

How to do this?

Use a variable called min to locate the index of the smallest elementmin 3


Selection SortAssume current iteration i = 1

2 8 6 4

0 1 2 3

Initialize min = i

min 1


Selection Sort Assume current iteration i = 1

Initialize min = I

ji

2 8 6 4

0 1 2 3

Initialize min = I Set j = i + 1Compare array(min) to array(j)

min 1


Selection Sort

j Assume current iteration i = 1Initialize min = iSet j = i + 1

i

2 8 6 4

0 1 2 3

Set j = i + 1Compare array(min) to array(j)If array(j) < array(min)

set min to j

min 2 Because 6 < 8,min is now set to 2min is now set to 2


Selection Sort

jIncrement j

i

2 8 6 4

0 1 2 3

Compare array(min) to array(j)

min 2


Selection Sort

j Increment jCompare array(min) to array(j)If array(j) < array(min)

i

2 8 6 4

0 1 2 3

If array(j) < array(min)set min to j

min 3

Because 4 < 6,min is now set to 3


Selection Sort

j

S (i) ith ( i )

i

2 8 6 4

0 1 2 3

Swap array(i) with array(min)

min 3


SortArrayvoid SortArray( double array[], int size){

i t i j iint i, j, min;

for (i=0; i < size‐1; i++) {{

min = i;for (j=i+1; j<size; j++) {{

if (array[j] < array[min]){

min = j;j}

}

Swap(&array[i], &array[min]); }

}


Swap void Swap (double *a, double *b) {{double temp = *a;*a = *b;*b = temp;

}


Swap void Swap (double *a, double *b) {{double temp = *a;*a = *b;*b = temp;

}Note: We’re passing two elements of the array; not passing the entire array

So, we CANNOT declare it as

void Swap(double a, double b)

void Swap(double a[], double b[])


Strings and File I/O


Character Type: char

In C, characters are indicated by single quotes:

The type is char for charmyChar;myChar = 'a';

f(“ h \ ” h )

ypone (1) character

printf(“myChar is %c\n”, myChar);

%c is the formal specifier for a char


String

A string is an array of characters, it is indicated by double quotes:

"this is a string"

But you can’t use an assignment operator to assign a string to a character arraychar myChar[80]; /* array of chars */myChar[4] = ‘a’; /* OK */

Ch “thi i t i ” /* NOT OK */myChar = “this is a string”; /* NOT OK */


String

A null (‘\0’) character is placed to mark the end of each string

t sisih a s \0gnirt

String functions use ‘\0’ to locate end of string (so you don’t have to pass string length as argument to the functions)


Initializing a Stringh [ ] {‘ ’ ’b’ ’ ’ ’d’ } myStrchar myStr[5] = {‘a’, ’b’, ’c’, ’d’, 0};printf(“myStr is %s\n”, myStr);

myStr

\0dcba

– Array size has exactly 5 elements– A terminating ‘\0’ character at the endA terminating \0 character at the end (in ASCII, char ‘\0’ is equivalent to int 0)

– Will print: “myStr is abcd”


Another waymyStr

char myStr[10]=“birdbath”;printf(“myStr is %s\n”, myStr);

b i r \0htabd

– Ok to use assignment only to initialize string– myStr is an array of 10 characters (but only the first 8 y y ( yelements used to store the word)

– a ‘\0’ is added to the end automatically (so 9 elements in th i d)the array are occupied)


Better waymyStr

char myStr[]=“birdbath”;printf(“myStr is %s\n”, myStr);

b i r \0htabd

– ‘\0’ is added automatically to the end of myStrS i ll t d t ti ll t h ld tl h t– Space is allocated automatically to hold exactly what we need (9 locations, including ‘\0’)


printf for Stringschar myStr[] = “abc123”;

Output:Output: – Use %s to print the entire string:

printf(“%s\n”,myStr); /* outputs abc123 *//* ‘\0’ is not printed */

– Use %c to print a character:printf(“%c\n”myStr[1]); /* outputs: b */printf( %c\n ,myStr[1]); /* outputs: b */


String input

There are several functions available

Part of the stdio.h library.


scanf

scanf reads up to the first white space, ignores the stuff typed in after that. Be careful when using it.

char myStr[10];i f(“E i “)printf(“Enter a string: “);

scanf(“%s”, myStr);printf(“You entered %s\n\n” myStr)printf( You entered %s\n\n , myStr)


scanf ‐ safer

Use %[width]s to copy only up to a maximum number of character. But, does not append the ‘\0’

char myStr[10];i f(“E i “)printf(“Enter a string: “);

scanf(“%9s”, myStr);myStr[9] = ‘\0’; /* Do this yourself just in case */myStr[9] = \0 ; / Do this yourself just in case / printf(“You entered %s\n\n”, myStr)


getchar

getchar will fetch one (1) character from the input stream

char myChar;printf(“Enter a character: “);

Ch h ()myChar = getchar();printf(“You entered %c\n\n”, myChar);


fgets

char * fgets(charArray, lengthLimit, filePtr)

• fetches a whole line, up to the size limit or when it sees a new line

• It will add a ‘\0’ at the end of string• Example• Example:

char myStr[80];fgets(myStr, 80, stdin); // fetch from console

• Returns a NULL if something went wrong, otherwise a pointer to the array

The functions that start with “f” work with any source of input. stdin is the C “Standard Input”.


p

#include<stdio.h>#include<string.h>

int main () {int inputLength=20int cnt=0;int cnt 0;char str1[20], str2[20];

printf("\nEnter a string: ");fgets(str1, inputLength, stdin);

Make sure you’re clear the difference between fgetsand scanfg ( , p g , );

printf("You entered %s\n",str1);

printf(“Enter another string: ");scanf("%s",str2);

and scanf

printf("You entered %s\n",str2);}


Note the extra new line when using fgets

str1 (read using fgets)

t sisih a s \ngnirt \0

str1 (read using fgets)

str2 (read using scanf)

t \0sih


String manipulation functions #include <string.h>

strcpy – Copies a stringstrcat – Concatenates two stringsstrlen – Returns the length of a stringstrcmp – Compares two strings


string copy (strcpy) strcpy(destString, srcString);

char [] strcpy (char dest[], const char src[]);

• copy string contents from src (2nd arg) to dest (1starg) including ‘\0’arg) including \0

• dest is changed, src unmodified (but can do some weird things)

• returns a pointer to the modified dest array


string copy (strcpy)

char [] strcpy (char dest[], const char src[]);

• There is no error checking!– If dest array is shorter than src array, no errors. Weird things could happen but no compile errors and often no runtimecould happen, but no compile errors and often no runtime errors

– Tracking these “bugs” down is very hard!g g y


Safer version of string copy strncpy(destString, srcString, 80);

char [] strncpy (char dest[], const char src[], int N);

• copies at most N characters from src to dest (or up to ‘\0’

• If length of src is greater than N, copies only the first N characters

• If length of src is less than N, pad the remaining elements in dest with ‘\0’

Does not copy the ‘\0’ if the string length is >= N


concatenation (strcat)

char [] strcat(char dest[], const char src[])

• contents of src are added to the end of dest. • dest is changed, src is not• ‘\0’ added to the end of dest• return a pointer to dest• no bounds check (again, this is C)


comparison (strcmp)

int strcmp (const char s1[],const char s2[]);

• Compares 2 strings– if s1 precedes s2, return value is less than 0– if s2 precedes s1, return value is greater than 0– if s1 equal to s2, return value is 0

C i i b d l i hi d (ASCII• Comparison is based on lexicographic order (ASCII order) – e.g., “a” < “b”


#include <stdio.h>#include <string.h>

int main() {printf("strcmp(\"a\",\"b\"): %d\n", strcmp("a","b"));printf("strcmp(\"b\" \"a\"): %d\n" strcmp("b" "a"));printf( strcmp(\ b\ ,\ a\ ): %d\n , strcmp( b , a ));printf(“strcmp(\"a\",\"a\"): %d\n", strcmp("a","a"));printf("strcmp(\"2\",\"3\"): %d\n", strcmp("2","3"));printf("strcmp(\"2\",\"10\"): %d\n", strcmp("2","10"));

}

Lexicographic ordering is not the same as ordering numbers


Reversing a string

Input: Math is funOutput: nuf si htaM

How would you do this?


front

backvoid Reverse(char str[] ){int front = 0; int back = strlen(str) ‐ 1;int back = strlen(str) ‐ 1; char t; /* A temporary place to put a character */

while (front < back){t = str[front];str[front] = str[back];str[back] = t;str[back] t;front++;back‐‐;

}}


}

2

Built‐in functions for characters#include <ctype.h>

Built in functions for characters

• These all return boolean (1/0). – isalnum(c): is c alphanumeric?– isalpha(c): is c alphabetic?– isdigit(c): is c a digit?– iscntrl(c): is c a control character?islower(c): is c lower case?– islower(c): is c lower case?

– isupper(c): is c upper case?– ispunct(c): is c a punctuation character, that is a printableispunct(c): is c a punctuation character, that is a printable character that is neither a space nor an alphanumeric character


converters

• char tolower(c)– return the character as lower case

• char toupper(c)– return the character as upper case


File IO

How to write programs to read from or write to filesor write to filesSo far, we’ve only looked at how to read/write to the consoleread/write to the console


Open files

• Files are opened with fopen– the fopen function is part of stdio.h

– It takes two args: a string (the name of a file) and a mode string (how the file is opened).string (how the file is opened).

– It returns a file pointer• fptr = fopen(“myfile.txt”,”r”);


Declare and assign file pointer

• fopen returns a NULL pointer if it cannot perform the operation

FILE *infile;inFile fopen(“file t t” ”r”)inFile = fopen(“file.txt”,”r”);if (inFile == NULL)// bad file opening


file modes

• “r”, read from the beginning of an existing file• “w”, make an empty file (wipe out old contents), start writing.

• “a”, find an existing file, start writing at the end of that file


Close files

• Files are closed with fclose– the fclose function is part of stdio.h

– It takes one argument: pointer to the file to be closedpointer to the file to be closed

fclose(fptr);


The “f” functions

• most of the I/O functions we know have an equivalent “f” version to work with a file.

• The first argument is the file pointer value of an opened file– fprintf(filePtr, “format string”, values)– fscanf(filePtr, “format string, addresses)f f(fil Pt ) d f fil t t– feof(filePtr) end of file test


States and State Machines


What is this for?

State machines are commonly used in…

Embedded Systems

Factory/Process Controls


State

State – An abstraction of the current status of a system. States are assigned names.

Waiting for a KeypressWaiting for Elvis

Paper JammedBattery is Below LimitWaiting for Elvis

Raising Firearm to FireCellphone is DialingD O i

Battery is Below LimitPower is OnDoor Open

Door Opening Prius Accelerator Stuck

Verbs with “ing” Statement of condition


States in a Garage Door Are there any more states?

DoorClosed

DoorOpen

CSE 251 Dr. Charles B. OwenProgramming in C4 A

More States

DoorOpening

DoorClosing


How we will express this in a program

/* Our possible garage door states */#define DoorClosed 1#define DoorOpening 2

int main(){i D Cl d#define DoorOpening 2

#define DoorOpen 3#define DoorClosing 4

int state = DoorClosed;…

In themain functionAbove main in our program

In the main function

Why do we care? We do different things y gdepending on the current state. What does the button do in each state?

CSE 251 Dr. Charles B. OwenProgramming in C6 B

Naming Conventions ‐ States

We will usually name states with camel‐case and a capital first letter. We’ll use #defines in our programs for state names.

WaitingForKeypressWaitingForElvis

PaperJammedl i iWaitingForElvis

RaisingFirearmCellphoneDialing

BatteryBelowLimitPowerOnDoorOpen

DoorOpeningp

PriusAccelStuck

Verbs with “ing” Statement of conditionVerbs with ing Statement of condition


EventsA i i i I iAn event is an occurrence in time. It is considered atomic and instantaneous.

Left mouse button pressedKey pressed

Paper is jammedMessage has timed outy p

Elvis has left the buildingFlight 529 has landedPower turned on

Message has timed out10 seconds has elapsedBattery is below limit

P j t 2 i dPower turned on Project 2 is due

Past tense verbs Onset of condition


Events vs. State

clickedOnScreen screenSlideDone clickedOnScreen screenSlideDone

IdleClosed IdleOpenSlidingOpen SlidingClosed IdleClosed

An event is what causes us to change from state to state.


State Diagrams State diagrams describe what we will implement in code as a state machine.


State Diagrams

Initial State

State

Transition

Event

Transition


Starting out State Machine

int main(){i D Cl dint state = DoorClosed;…

CSE 251 Dr. Charles B. OwenProgramming in C12 C

A Control Loopint main(){int state = DoorClosed;int state = DoorClosed;

printf("Garage Startup\n");GarageStartup();

A continuous loop in a controls application that controls the

t Ri ht l iwhile(IsGarageRunning()){

system. Right now our loop is doing nothing. We’ll want to add code to make our garage door

}

printf("Garage Shutdown\n");G Sh d ()

work.

GarageShutdown();return 0;

}


Important Idea

We do something different depending on the state we are in. It makes sense to create a function for each state.

void StateDoorClosed(int *state){ Note the pointer. We

th t t b} pass the state by reference. It is an in/out parameter

What should happen when we are in the DoorClosed state?


DoorClosed state…

If the button is pressed:Start the motorGo to the DoorOpening statep g

otherwise:Do nothing…


DoorClosed state…id St t D Cl d(i t * t t )

If the button is pressed:Start the motorGo to the DoorOpening state

void StateDoorClosed(int *state){if(WasButtonPressed()){p g

otherwise:Do nothing…

SetMotorPower(1);*state = DoorOpening;

}}}


The Control Loop – Handling Stateswhile(IsGarageRunning()){switch(state){

We will put a switch statement in our control

{case DoorClosed:StateDoorClosed(&state);break;

loop to handle the states.

}

}


We now have this…

Trigger / Activity

Trigger is what causes the transition Activitythe transition. Activity is something that happens when the transition occurs.

This is a simple 2‐state statement.


What happenswhile(IsGarageRunning()){switch(state){ void StateDoorClosed(int *state){case DoorClosed:StateDoorClosed(&state);break;

void StateDoorClosed(int state){if(WasButtonPressed()){

}

}

SetMotorPower(1);*state = DoorOpening;

}}

Control LoopState Function}

The control loop runs continuously (1000 times per second in this program). Each time it calls a function for the current state. That state function decides if we need to change the state and does anything else we need to do while in that state.


A Garage Door Opener


Pointers, Arrays, Multidimensional Arrays

• Pointers versus arrays– Lots of similarities

• How to deal with 2D, 3D, multidimensional arrays (for storing matrices and other 2D or 3D data!)(for storing matrices and other 2D or 3D data!)


Review: PointersAddress Memory Name

Pointers are variables that store memory addresses

Address Memory

0xeffffa94

Name

a15

int a = 15;

0xeffffa98 b0xeffffa94

int a 15;int *b = &a;

printf(“%x %x %d\n”, b, &b, *b);// prints effffa94 effffa98 15// prints effffa94 effffa98 15


int number; int *ptr = &number;

Using pointers in scanf function

Read & as “at”

printf(“Enter an integer: ”);scanf(“%d” &number);

scanf functionas at

scanf( %d , &number);printf(“Enter another integer: “);scanf(“%d”, ptr);

Don’t have to put & before ptr. It’s already an “at”

printf(“Number = %d, *ptr = %d\n”, number, *ptr);

already an at .

Example output:


int multiply( int *, int);

int main()

Passing pointers (addresses) toint main()

{int number = 3;int *ptr &n mber;

(addresses) to functions

int *ptr = &number;printf(“1: %d\n”, multiply( &number, 2 ) );printf(“2: %d\n”, multiply( ptr, 3 ) );

}}

int multiply (int *a, int factor) {{

return (*a) * factor;}


Review: Arrays

An array is a contiguous chunk of memory to store multiple values

int grades[]={74,59,95,85,71,45,99,82,76};

0xeffffa00 0xeffffa04 0xeffffa08 0xeffffa0c 0xeffffa10 0xeffffa14 0xeffffa18 0xeffffa1c 0xeffffa20

grades 74 59 95 85 71 45 99 82 76

index 0 1 2 3 4 5 6 7 8


int sumArray( int [], int);

int main() Passing arrays to functions{

int list[] = {1, 2, 3, 4};printf(“Sum = %d\n”, sumArray( list , 4 ));

functions

}

int sumArray (int list[], int arraySize) {{

int sumvalue = 0;for (int i=0; i<arraySize; i++)

sumvalue += list[i];sumvalue += list[i];return sumvalue;

}


Array Nameffe2de0c ffe2de10 ffe2de14 ffe2de18The array name is a

pointer to the first l t f th

1 2 3 4

[1] [2] [3][0]

listffe2de0c ffe2de10 ffe2de14 ffe2de18

element of the array [1] [2] [3][0]

int list[]={1,2,3,4};

Output: ffe2de0c ffe2de0c 1

int list[] {1,2,3,4};printf(“%x, %x, %d”, list, &list[0], *list);

Output: ffe2de0c ffe2de0c 1


Pointers and Arrays p

int *p, int list[] {1 2 3 4};

1 2 3 4

[1] [2] [3][0]

list

int list[]={1,2,3,4}; p = list; /* equivalent to p = &list[0] */printf(“%d\n”, *p); /* prints the value “1” */

[ ] [ ] [ ][ ]

YYou can use apointer to access

the array


Pointer and [] P

Any pointer to a block of memory can use the [] syntax, even if it is

not declared as an array! 1 2 3 4Listnot declared as an array!

[1] [2] [3][0]

int *p, pint list[]={1,2,3,4}; p = list;i tf(“%d\ ” [2]) // i t 3

int *v; and int v[]; /* Mean the same thing */

printf(“%d\n”, p[2]); // prints 3


Array indexing [] *list – Contents pointed to by list*(list + 2) – Contents at list[2]

Indexing an array is just a way

list

ff 2d 0 ff 2d 10 ff 2d 14 ff 2d 18

list + 2

of finding a particular address in that block1 2 3 4

[1] [2] [3][0]

ffe2de0c ffe2de10 ffe2de14 ffe2de18

[1] [2] [3][0]

int list[] = {1 2 3 4} // array of 4 intsint list[] = {1,2,3,4} // array of 4 intsprintf(“%d”, list[2]);

This is equivalent toprintf(“%d”,*(list+2));printf( %d , (list+2));


Pointer Arithmetic

When we add to a pointer, such as (p + 1), we don’t literally add 1 to the pointer address

Instead we add one “address” to the pointer


Pointer Arithmeticint list[] = {1, 2, 3, 4};int *p = list; /* same as p = &list[0] */printf(“%x”,p); /* prints ffe2de0c */

p

printf( %x ,p); / prints ffe2de0c /


1 2 3 4


Pointer Arithmeticint list[] = {1, 2, 3, 4};int *p = list; /* same as p = &list[0] */printf(“%x”,p); /* prints ffe2de0c */printf( %x ,p); / prints ffe2de0c /p = p + 1; /* p increases by 4 */printf(“%x”,p); /* prints ffe2de10 */

pThink of pointer arithmetic as add

1 2 3 4

ffe2de0c ffe2de10 ffe2de14 ffe2de181 “location” instead of one byte or addressbyte or address.


Pointer Arithmeticd bl li t2[] {1 0 2 0 3 0}double list2[] = {1.0, 2.0, 3.0};double *p = list2; /* same as p = &list2[0] */printf(“%x”, p); /* prints ffe2de0c */

p

p ( p) p

1.0


2.0 3.0

ffe2de1c ffe2de20

1.0 2.0 3.0


Pointer Arithmeticd bl li t2[] {1 0 2 0 3 0}double list2[] = {1.0, 2.0, 3.0};double *p = list2; /* same as p = &list2[0] */printf(“%x”,p); /* prints ffe2de0c */p ( p) pp = p + 1; /* P increases by 8 bytes */printf(“%x”,p); /* prints ffe2de14 */

P

1.0


2.0 3.0

ffe2de1c ffe2de20


Pointer Arithmetic on Arrays*(l ) f h l h• *(list+1) references the next element in the array (equivalent to list[1])

• Be careful: *(++list) works too but now we have lost our pointer to the beginning of the array!!!our pointer to the beginning of the array!!!– Equivalent to: list = list + 1; *list;


sizeof() operatori t i

Returns the number of bytes needed toint i;

int *ptr4i = &i;int IntArray[] = {1, 2, 3, 4, 5};double j;

of bytes needed to store a variable or a data typedouble j;

double *ptr4j = &j;double doubleArray[] = {1.0, 2.0, 3.0, 4.0,5.0};

i tf("Si f i t i %d b t \ " i f(i t))

data type

printf("Sizeof integer is %d bytes\n", sizeof(int));printf("Sizeof double is %d bytes\n", sizeof(double));

printf("Sizeof i is %d bytes\n", sizeof(i));printf("Sizeof pointer for i is %d bytes\n", sizeof(ptr4i));

printf("Sizeof j is %d bytes\n", sizeof(j));printf("Sizeof pointer for j is %d bytes\n", sizeof(ptr4j));

printf("Sizeof intArray is %d bytes\n", sizeof(intArray));printf("Sizeof doubleArray is %d bytes\n", sizeof(doubleArray));


sizeof() operator


When we pass an arrayWh i thWhen we pass an array, we are passing the array address

int sumArray( int [ ] int);int sumArray( int [ ], int);

int main() {{int list[] = {1, 2,3, 4); …

A ( li 4 )sumArray( list , 4 );

…


When we pass an array

This will work too (because array name is a pointer to the beginning of the array)

int sumArray( int *, int);int main() {{int list[] = {1, 2,3, 4); …sumArray( list , 4 );

…


When we pass an array

• But this DOES NOT work!

int sumArray( int *, int);y( , );int main() {int *list {1 2 3 4);int *list = {1, 2, 3, 4); …sumArray( list , 4 );

…


Pointers *list – Contents pointed to by list*(list + 2) – Contents at list[2]

Indexing an array is just a

list

ff 2d 0 ff 2d 10 ff 2d 14 ff 2d 18

list + 2

way of finding a particular address in that block1 2 3 4

[1] [2] [3][0]


[1] [2] [3][0]

int list[] = {1 2 3 4} // array of 4 intsint list[] = {1,2,3,4} // array of 4 intsprintf(“%d”, list[2]);

This is equivalent toprintf(“%d”,*(list+2));printf( %d , (list+2));


2‐D Arrays

int cave[ArraySize][ArraySize];

Column

1 2 3 40

0 1 2 3

Column

1 2 3 45 6 7 89 10 11 12

0

1

2Row

9 10 11 1213 14 15 16

2

3


2D Arraysint myMatrix[3][4] = { {1 2 3 4} {5 6 7 8} {9 10 11 12} };int myMatrix[3][4] = { {1,2,3,4},{5,6,7,8},{9,10,11,12} };

0 1 2 3

Column

1 2 3 45 6 7 8

myMatrix[0][1] → 2

0 1 2 3

0

1Row 5 6 7 89 10 11 12

M t i [2][3] 12

2

Row

myMatrix[2][3] → 12myMatrix[row][col]


Physically, in one block of memoryint myMatrix[2][4] = { {1,2,3,4},{5,6,7,8} };

ffe2de0c ffe2de10 ffe2de14 ffe2de18 ffe2de1c ffe2de20 ffe2de24 ffe2de28

1 2 3 4 5 6 7 8ffe2de0c ffe2de10 ffe2de14 ffe2de18 ffe2de1c ffe2de20 ffe2de24 ffe2de28

Array elements are stored in row major order

row 1 row 2

Array elements are stored in row major orderRow 1 first, followed by row2, row3, and so on


2D Array Name and Addresses[ ][ ] { { } { } }

ffe2de0c ffe2de10 ffe2de14 ffe2de18 ffe2de1c ffe2de20 ffe2de24 ffe2de28

int myMatrix[2][4] = { {1,2,3,4},{5,6,7,8} };

1 2 3 4 5 6 7 8ffe2de0c ffe2de10 ffe2de14 ffe2de18 ffe2de1c ffe2de20 ffe2de24 ffe2de28

myMatrix: pointer to the first element of the 2D arraymyMatrix[0]: pointer to the first row of the 2D arraymyMatrix[1]: pointer to the second row of the 2D array*myMatrix[1] is the address of element myMatrix[1][0]


Accessing 2D Array Elements[ ][ ] { { } { } }int myMatrix[2][4] = { {1,2,3,4} , {5,6,7,8} };

1 2 3 4 5 6 7 81 2 3 4 5 6 7 8

Indexing: myMatrix[i][j] is same asg y [ ][j]*(myMatrix[i] + j)(*(myMatrix + i))[j]*((*(myMatrix + i)) + j)*((*(myMatrix + i)) + j)*(&myMatrix[0][0] + 4*i + j)


Declaration#define ROWS 3#define COLS 5

int table[ROWS][COLS];

void display (table);


void display( int x[ROWS][COLS] ) {

2D Arrays often require nested loops – two

for (int i=0; i < ROWS; i++) {

for (int j=0; j < COLS; j++ )

pvariables

( j ; j ; j ){

printf(" x[%d][%d]: %d", i, j, x[i][j]);}}printf("\n");

}printf("\n");

}


Table A = { {13, 22, 9, 23},{17, 5, 24, 31, 55},{4 19 29 41 61} };

13 22 9 23 ?17 5 24 31 55{4, 19, 29, 41, 61} };4 19 29 41 61

Table B = {1, 2, 3, 4,5, 6, 7, 8, 9, 10 11 12 13 14 };

1 2 3 4 56 7 8 9 1011 12 13 14 ?10, 11, 12, 13, 14 }; 11 12 13 14 ?


passing 2d arrays

In passing a multi‐dimensional array, the first array size does not have to be specified. The second (and any subsequent) dimensions must be given!

int myFun(int list[][10]);


#define ROWS 3#define COLS 5

int addMatrix( int [ ][COLS] );int addMatrix( int [ ][COLS] );

int main() {int a[][COLS] = { {13 22 9 23 12} {17 5 24 31 55} {4 19 29 41 61} };int a[][COLS] = { {13, 22, 9, 23, 12}, {17, 5, 24, 31, 55}, {4, 19, 29, 41, 61} };printf(“Sum = %d\n”, addMatrix( a ) );

}

int addMatrix( int t[ ][COLS] )int addMatrix( int t[ ][COLS] ) {int i, j, sum = 0;for (i=0; i<ROWS; i++)for (i=0; i<ROWS; i++) for (j=0; j<COLS; j++) sum += t[i][j];

return sum;;}


Compilation and Makefiles


Lecture Outline• What is gcc?

– What are the different switches available?

• What is a Makefile?

• Exercise:– Take an existing file, split it into multiple separateTake an existing file, split it into multiple separate programs, and write a Makefile to simplify compilation of the files


gcc is not a single program

gcc is a conglomeration of programs that work together to

make an executable filemake an executable file

We need to know what it does and how we can better control it.and how we can better control it.

In particular, we would like to be able to make multiple, separate p , p

programs which can be combined into one executable


What does gcc do?#i l d tdi hd #include <stdio.h>#include <math.h>

#define NumIter 5

code.c

int main() {int i;

C sourcefile

int i;for (i=1; i<=NumIter; i++)

printf(“PI^%d is %f\n”, i, pow(M_PI, i));}


What does gcc do? ExpandedC code

C ilpreprocessor(# commands)

Compiler

C sourceMachine Code( o file)C source

file(.o file)

Linker

Executable

LibraryFilesFiles


Some gcc optionsP i ( d d i)• Preprocessing (gcc –E code.c > code.i)– Removes preprocessor directives (commands that start with #)– Produces code.i Don’t use directlyProduces code.i Don t use directly

• Compiling (gcc –o code.o –c code.i)– Converts source code to machine language with unresolved directives– Produces the code.o binary

• Linking (gcc –lm –o code code.o)– Creates machine language exectutable– Produces the code binary by linking with the math library (‐lm)


C Preprocessor, cppExpanded

Processes commands that are preceded with # symbols and

preprocessor(# commands)

ExpandedC code

p yexpands the code

#include(# commands)

#define#ifdef and #ifndef

C sourcefile


#include

Include header files that contain declarations necessary for compiling code

– the declarations provide the proper types, but not the actual definitions (the actual code)


data c

# ifndef _STDIO_H

# if !defined need FILE &&

stdio.h

#include <stdio.h>

#include “functions.h”

data.c __ _!defined __need___FILE

# define _STDIO_H 1

# include <features.h>

int main() {

double x = 1;

__BEGIN_DECLS

…

printf(“Sinc(x) = %f\n”, Sinc(x));

} double Sinc(double x);

functions.h

gcc –E data.c > data.idata i is like an “expanded” C code in which the #include commanddata.i is like an expanded C code in which the #include command is “substituted” with text from the files


Example#include “display.h”display h

main.c

void DisplayMe(int n); int main(){

DisplayMe(12);

display.h

DisplayMe(12);}


Header Files

We’ve been using .h files to define things from the C libraries:stdio.h, stdlib.h, stdbool.h, etc.

These are called Header Files.

We’re going to create our own .h files to share information between .c files.


#defined f bl#defines a preprocessor variable

– every place the variable occurs, the definition will be substituted as codesubstituted as code

Syntax: #define var_name var_definitiony _ _– Examples:

#define DEBUG 1#d fi PI 3 1415926535#define PI 3.1415926535


#include <stdio.h>#define PI 3 14157265

preproc.c

gcc –E preproc c > preproc i#define PI 3.14157265#define square(x) (x*x)

int main() {

gcc E preproc.c > preproc.i

printf("Value of PI is %f\n", PI);printf("Square of 3.5 is %f\n", square(3.5));

}

# 1 “preproc.c"# 1 "<built‐in>"# 1 "<command line>“

preproc.i

# 1 <command‐line>…typedef unsigned int size_t;…int main() {printf("Value of PI is %f\n", 3.14157265);printf("Square of 3.5 is %f\n", (3.5*3.5));

}


}

#ifdef, #ifndef, #ifSurround code that might be included. Include for compilation when desired

#ifdef DEBUG….#endif#ifndef DEBUG#ifndef DEBUG…#endif#endif


Common useif(num == 1){/* This is the only time we actually move a disk */DisplayTower(tower);DisplayTower(tower);

#if 0printf("Press return");

( ( ) )

This is a handy way to turn off a block of code without removing it It often worksfgets(cmd, sizeof(cmd), stdin);

#endif

MoveDisk(tower, fm, to);

removing it. It often works better than trying to comment out the block, MoveDisk(tower, fm, to);

return;}

which may have comments already in it.


Header Files #include <stdio.h>#include <stdlib.h>hanoi h:

hanoi.c:

#define NumDisks 6#define NumPins 3

#include <stdbool.h>

#include “hanoi.h”

hanoi.h:

bool CheckDone(int tower[NumPins][NumDisks]);

#i l d di h

display.c:I have two .c files in the same program. B h d k #include <stdio.h>

#include <stdlib.h>#include <stdbool.h>

Both need to know NumDisks and NumPins. So, I put that


void DisplayTower(int tower[NumPins][NumDisks]){

information in a header file: hanoi.h

{…}


Example#include “hanoi.h”hanoi h:

hanoi.c:



#if 0

hanoi.h:

#if 0bool DebugStuff(int x, int b[NumPins]);#endif


What would happen? #include <stdio.h>#include <stdlib.h>hanoi h:

hanoi.c:



#include “solve.h”#include “hanoi h”

hanoi.h:

#include hanoi.h

bool CheckDone(int tower[NumPins][NumDisks]);#include “hanoi.h”

solve.h:

void Autosolve(inttower[NumPins][NumDisks]);


Include Guards #include <stdio.h>#include <stdlib.h>hanoi h:

hanoi.c:

#ifndef HANOI_H#define HANOI_H



hanoi.h:


# dif

#include hanoi.h


#endif

#ifndef SOLVE H

solve.h: Include Guards: Conditional compilation code that protects a #ifndef SOLVE_H

#define SOLVE_H


section of code in a header from being compiled more than once.



#endif

Compiling ExpandedC code

Object Code (.o file)

gcc –c code.c

preprocessor(# commands)

Compiler

C

Compiler translates the C source code into object code – what the machine actually

C sourcefile understands

Machine‐specific

Each line represents either a piece of data orEach line represents either a piece of data or a machine‐level instruction

To create the assembly code:gcc -c preproc.c


Linking• Object file may not be directly executable

– Missing some partsStill has some names to be resolved– Still has some names to be resolved

• The linker (ld) takes multiple object files ( o) and putsThe linker (ld) takes multiple object files (.o) and puts them together into one executable file– Resolves references to calls from one file to another

• Linking is important as it allows multiple, separate fil t b b ht t thfiles to be brought together


Linkingpreprocessor

ExpandedC code

Compilergcc ‐E

p

gcc ‐c

C source

Machine Code(.o file)

file

Linker

ExecutableExecutable

LibraryFiles

gcc –lm code.o –o code


Separate compilation

• For large, complex programs, it is a good idea to break your files into separate .c and .h files

• Can debug each separately, then reuse

• Can distribute pieces to other so they can incorporate into their code without changing their code (just link it in)


Example (Separate Compilation)shapes.h shapes.cmainprog.c

math.h

mainprog

gcc -c shapes.cg p

gcc –c mainprog.c

gcc –lm mainprog.o shapes.o –o mainprog


Another way (Compile all at once)shapes.h shapes.cmainprog.c

math.h

mainprog

l h i igcc –lm shapes.c mainprog.c –o mainprog


Controlling compilation/linking better

• We need a better way to control compilation– too much detail to type each time

• We can save time by only updating (compiling) the parts that need updating– if a .o file is OK (unchanged source), leave it. If a .c file is changed, the .o file needs to be regenerated then everything relinkedeverything relinked


Makefiles

• Suppose you have a program, which is made up of these 5 source files:main.c (include data.h, io.h)data.c (include data.h)d hdata.hio.c (include io.h)io hio.h


How to create the executable?

data .c data.h main.c io.cio.h

gcc -c data.c → data.ogcc c io c → io ogcc -c io.c → io.ogcc -c main.c → main.o

i i d t t blgcc main.o io.o data.o –o executable


How to create the executable?

data .c data.h main.c io.cio.h

gcc -c data.cgcc c io c

What if you modify data.h?

D d tgcc -c io.cgcc -c main.c

i i d t t bl

Do you need to re‐run gcc ‐c on io.c and main.c?

gcc main.o io.o data.o –o executable


Dependencies executable

data.o main.o io.o

data.c data .h main.c io.cio.h


What is a Makefile?

• A Makefile is used with the make utility to determine which portions of a program to compile.

• It is basically a script that guides the make utility to choose the appropriate program files that are to be compiled and linked together


If data.h is modified executable

d t i idata.o main.o io.o

data .c data .h main.c io.cio.h

No need to re‐create data.o and main.o


If io.c is modified executable

d t i idata.o main.o io.o


No need to re‐create data.o and main.o


#ifndef HANOI_H#define HANOI_H

#include <stdio.h>#include <stdlib.h>

hanoi.h: hanoi.c:




#endif#include hanoi.h

bool CheckDone(int tower[NumPins][NumDisks]);solve.h:

#ifndef SOLVE_H#define SOLVE_H

#include “hanoi h”

#include <stdio.h>#include <stdlib.h>

display.c:

#include hanoi.h




#endifbool DisplayTower(int tower[NumPins][NumDisks]);

CSE 251 Dr. Charles B. OwenProgramming in C34 D

What is in a Makefile?

name incolumn1 spaces

fil d d 1 d d 2

not commas

file: dependency1 dependency2command

TAB character


What it means

• The first line indicates what a file depends on. That is, if any of the dependencies change, then that file must be updated– What updating means is defined in the command listed below the dependencybelow the dependency

• Don’t forget the TAB character to begin the second• Don t forget the TAB character to begin the second line


Dependency graph Executable (prog)

data.o main.o io.o



Creating a Makefile Makefile:

Executable (prog) prog:

data.o:

data.o main.o io.o

main.o:

io.o:


List the object (*.o) and executable files that make up the program


Creating a Makefile Makefile:

d t i iExecutable (prog) prog: data.o main.o io.ogcc –o prog data.o main.o io.o

data.o: data.h data.c

data.o main.o io.ogcc –c data.c

main.o: data.h io.h main.cigcc –c main.c

io.o: io.h io.cgcc –c io.c


First line specifies the dependencies for

g

each object and executable files


Creating a Makefiled t i i

Makefile:

Executable (prog)prog: data.o main.o io.o

gcc –o prog data.o main.o io.o

data.o: data.h data.c

data.o main.o io.ogcc –c data.c

main.o: data.h io.h main.cigcc –c main.c

io.o: io.h io.cgcc –c io.c

data .c data .h main.c io.cio.hg

Second line contains the command to execute if any of the dependence files y pare modified


Additional makefile “targets”

clean: rm -rf *.o example22

Command “make clean” calls the clean commandIn this case clears the o files and the executableIn this case, clears the .o files and the executable


Usage

• make– To create the executable

• make linkedList.o– do what has to be done to make that .o file, then stop

• make clean– run the clean command


structs Box

Aggregating associated data into a single variable

widthlengthheightheight

Circleint main() {

radiusBox mybox; Circle c;

mybox width = 10;mybox.width = 10;mybox.length = 30;mybox.height = 10;c.radius = 10;

}


The idea Boxwidthlengthheight

I want to describe a box. I need variables for the width, length, and height. height

I can use three variables, but wouldn’t it be better if I had a single variable to describe a better if I had a single variable to describe abox?

That variable can have three parts the widthThat variable can have three parts, the width, length, and height.


Structs

A struct (short for structure) in C is a grouping of variables together into a single type

– Similar to structs in Matlab

struct nameOfStruct{

type member;type member;…

}; N t th i l t th d}; Note the semicolon at the end.To declare a variable:

struct nameOfStruct variable_name;


Example #include <stdio.h>

struct Box

Boxidth

struct Box {

int width;int length;

Data structure

widthlengthheight

int length;int height;

};

l

definition

Circle

struct Circle {

double radius;Circle

radius};

int main() {

You can declare

i bl{struct Box b; struct Circle c;

}

variables


}

A

Example

int main() {struct Box b; You can

assign values

#include <stdio.h>

struct Box {

b.width = 10;b.length = 30;b.height = 10;

gto each member

int width;int length;int height;

};

Box

b.height 10;

}

};

widthlengthheight

We use a period “.” to get to the elements of a struct. g

If x is a struct, x.width is an element in a struct


in a struct.

Another Examplet t b kR dSt t

You can use mixed data types

struct bankRecordStruct{

char name[50]; within the struct (int, float, char [])

char name[50];float balance;

};};

struct bankRecordStruct billsAcc;


Accessing valuest t b kR dSt tstruct bankRecordStruct{

char name[50];Access values in a struct using a period:

float balance;};

“.”

struct bankRecordStruct billsAcc;

i tf(“M b l i %f\ ” bill A b l )printf(“My balance is: %f\n”, billsAcc.balance);

float bal = billsAcc.balance;


Assign Values using Scanf()t t k dstruct BankRecord

{char name[50];float balance;;

};

int main() (){

struct BankRecord newAcc; /* create new bank record */

i tf(“E t t “)printf(“Enter account name: “);scanf(“%50s”, newAcc.name);printf(“Enter account balance: “);scanf(“%d”, &newAcc.balance);scanf( %d , &newAcc.balance);

}


Copy via =

You can set two struct type variables equal to each other and each element will be copied

struct Box { int width, length, height; };

int main() {

struct Box b, c;b.width = 5; b.length=1; b.height = 2;c = b; // copies all elements of b to cc = b; // copies all elements of b to cprintf(“%d %d %d\n”, c.width, c.length, c.height);

}


Passing Struct to a function

• You can pass a struct to a function. All the elements are copied

• If an element is a pointer, the pointer is copied but not what it points to!

int myFunction(struct Person p){…}


Using Structs in Functions

Write a program that – Prompts the user to enter the dimensions of a 3D box and a circle

– Prints the volume of the box and area of the circle

Sample run:


#include <stdio.h>#include <math.h>

struct Box { int width height length; };struct Box { int width, height , length; };

int GetVolume(struct Box b) {

return b.width * b.height * b.length;}


struct Box b;

printf("Enter the box dimensions (width length height): ");scanf("%d %d %d", &b.width, &b.length, &b.height);

printf("Box volume = %d\n", GetVolume(b));printf( Box volume %d\n , GetVolume(b));}


Note: == Comparison doesn’t workstruct Box { int width, length, height; };


struct Box b, c;b width 5; b length 1; b height 2;b.width = 5; b.length=1; b.height = 2;c = b;if (c == b) /* Error when you compile! */

printf(“c and b are identical\n”);else

printf(“c and b are different\n”);printf( c and b are different\n );} t

Error message: invalid operands to binary == (have 'Box' and 'Box')


Error message: invalid operands to binary == (have Box and Box )

Create your own equality test#i l d < tdi h>#include <stdio.h>#include <math.h>

struct Box { int width, height , length; };{ , g , g ; };

int IsEqual(struct Box b, struct Box c) {

if (b width==c width &&if (b.width==c.width &&b.length==c.length &&b.height==c.height)return 1;

belsereturn 0;

}

struct Box b, c;b.width = 5; b.length=1; b.height = 2;c = b;

if (IsEqual(b,c))printf("c and b are identical\n");

elsei tf(" d b diff t\ ")


printf("c and b are different\n");

D

typedef

typedef is a way in C to give a name to a custom type.

typedef type newname;

typedef int dollars;typedef int dollars;typedef unsigned char Byte;

I can declare variables like:

dollars d;Byte b c;Byte b, c;

It’s as if the type already existed.


typedef for Arrays

There is a special syntax for arrays:Now, instead of:

typedef char Names[40];typedef double Vector[4];typedef double Mat4x4[4][4];

double mat[4][4];

typedef double Mat4x4[4][4];I can do:

Mat4x4 mat;Mat4x4 mat;


Using Structs with Typedef

typedef struct [nameOfStruct] {

type member;type member; optionalyp…

} TypeName;

optional

} yp ;

To declare a variable: TypeName variable_name;


Example#include <stdio.h>

typedef struct{Box

widthl th

{int width;int length;i t h i htlength

height

Ci l

int height;} Box;

typedef struct { double radius; } Circle;Circle

radius

yp { }

int main() {

Box b; /* instead of struct Box */Box b; /* instead of struct Box */Circle c; /* instead of struct Circle */b.width = 10;b l th 30b.length = 30;b.height = 10;c.radius = 10;

}


}

E

Arrays of structs

You can declare an array of a structure and manipulate each one

typedef struct{

d bl didouble radius;int x;int y;h [10]char name[10];

} Circle;

Circle circles[5];


Size of a Struct: sizeof

typedef struct{

double radius; /* 8 bytes */int x; /* 4 bytes */int y; /* 4 bytes */int y; /* 4 bytes */char name[10]; /* 10 bytes */

} Circle;} Circle;

printf(“Size of Circle struct is %d\n”, sizeof(Circle));sizeof(Circle));


Size of a Struct8 + 4 + 4 + 10 = 26

– But sizeof() reports 28 bytes!!!

Most machines require alignment on 4‐byte boundary ( d)(a word)

– last word is not filled by the char (2 bytes used, 2 left over)

D D D D D D D D I I I I I I I I C C C C C C C C C C X X8 byte, 2 word double 4 byte,

1 word4 byte, 1 word

10 byte char array, 2 bytesof the last word unused

integer integerof the last word unused


Pointers to structs

typedef struct{int width;

Box b; /* A variable of type Box */Box *c;/* A pointer to a Box */double w;int length;

int height;} Box;

double w;

b.width = 5; b.height = 7; b.length = 3;

c = &b; /* Same as before */

w = c‐>width;

To access the members of a struct, we use:

. for a variable of the struct’s type‐> for a pointer to a struct


> for a pointer to a struct

struct Concepts struct Box b; /* No typedef */Circle c; /* typedef */

struct Box{

double wid, hit;}

struct Box *pBox; /* Pointer to Box */Circle *pCirc; /* Pointer to Circle */

};

typedef struct{

pBox = &b; /* Get pointer to a Box */b.wid = 3;pBox‐>wid = 7;{

double radius;int x;int y;char name[10];

pBox >wid 7;

pCirc = &c;(*pCirc).radius = 9;char name[10];

} Circle;


Dynamic Memory Allocation

• Dynamic memory allocation– How to allocate memory for variables (esp. arrays/strings) during run timeduring run time

– malloc(), calloc(), realloc(), and free()


Why dynamic memory allocation? Usually, so far, the arrays and strings we’re using have fixed length (i.e., length is known at compile time)

• Example:h [ ] // ll f hchar myStr[11]; // allocates memory for 10 charsprintf(“Enter a string: “);fgets(myStr, 11, stdin);fgets(myStr, 11, stdin);

What if the user wants to enter a string more than 10enter a string more than 10 chars long or if the length is known only at run time?


y

malloc()

• malloc() is used to request additional memory from the operating system during program execution

Syntax: malloc(numBytes)

• Input is the number of consecutive bytes to be allocated• Return value is a pointer to the beginning of the block of

memory allocated or NULL if malloc fails

• To use malloc(), you must #include <stdlib.h>


malloc()h * h P /* d l i h */char *charP; /* declare a pointer to char */

charP

charP = malloc(10); 10 bytes or charscharP malloc(10);

charP

10 bytes or chars

charP contains the address of the beginning of that block.


free()

• The function free() returns memory to the memory pool. It “frees” up memory

Syntax: free(ptr)

where ptr “points to” memory previously allocated by– where ptr points to memory previously allocated by malloc() function

• To use free(), you must #include <stdlib.h>


Example

This program allows the user to specify the length of a stringspecify the length of a string, allocates memory for the string, and then applies string operationspp g p


#include <stdlib.h> /* you need this library for malloc(), free(), exit() */#include <stdio.h>#include <string.h> /* you need this library for strchr(), strlen() */

int main () {char *charP, *q;, q;int maxlen;

printf("Enter maximum string length: ");scanf("%d" &maxlen);scanf( %d , &maxlen);getchar(); /* reads the newline character */

printf("Enter the string: ");p ( g )fgets(charP, maxlen, stdin);

if ((q = strchr(charP, '\n')) != NULL) *q = '\0';

printf("You've entered a string %s of length %d\n", charP, strlen(charP));free(charP);}


}

A

What it does:

if ((q = strchr(charP, '\n')) != NULL) *q = '\0';

The function fgets returns the entire line input by the user including the newline at the end (when the user hit return). The function strchr(charP, ‘\n’) will return a pointer to that character (newline) if it exists or NULL otherwise Ifwill return a pointer to that character (newline) if it exists or NULL otherwise. If the result in the variable q is not NULL, the if statement executes the line of code to replace the newline with a null termination for the string.


strchr

if ((q = strchr(charP, '\n')) != NULL)*q = '\0';

This code finds the first occurance of the character ‘\n’ which is the newline character that fgets will obtain If found (value in q isthat fgets will obtain. If found (value in q is not NULL), it sets that character to the string null termination.


Memory leak

If malloc’ed memory is not free’ed, then the OS will “leak memory”

– This means that memory is allocated to the program but not returned to the OS when it is finished using itThe program therefore grows larger over time– The program therefore grows larger over time.


Exampleint main () {char *charP, r[80]; Memory leak exampleint length;

while (1) {{

printf("Enter the maximum string length: ");fgets(r, 80, stdin);sscanf(r, "%d", &length);

if ((charP = malloc(length)) == NULL) {printf("Out of memory. Quitting\n");exit(1);( );

}printf("Enter the string: ");fgets(charP, length, stdin);/* free(charP); */

}}


Exampleint main () {char *charP, r[80];

Memory leak exampleEach iteration of the loop allocates memoryint length;

while (1) {

Each iteration of the loop allocates memory for a string. But, that memory is never freed. Hence, we have a memory leak.{

printf("Enter the maximum string length: ");fgets(r, 80, stdin);sscanf(r, "%d", &length);

if ((charP = malloc(length)) == NULL) {printf("Out of memory. Quitting\n");exit(1);( );

}printf("Enter the string: ");fgets(charP, length, stdin);/* free(charP); */

}}


malloc without freeing while (1) {

charP = malloc(length+1);charP = malloc(length+1);…

}

charP

If you don’t free the allocated memory, previous block is still “ours” according to the OS, but we can no longer find it (no pointer to it). That block is an orphan!

It’s like you bought a house, but then lost the address You still own it and pay taxes on it butaddress. You still own it and pay taxes on it, but you can’t use it because you can’t find it.


Always free what you malloc

• You are responsible for your memory, you must allocate it and free it.

• Unlike other languages, it is all up to you!

• If you don’t free it, your program grows larger and eventually runs out of memory!


malloc allocates bytes

• If you want a character array that stores 10 characters (including ‘\0’):

char *p = malloc(10);

If t t ll t t f 10 i t ( d bl• If you want to allocate storage for 10 ints (or doubles or floats), you can’t do this:

int *p = malloc(10); /* WRONG! Why? */int p = malloc(10); / WRONG! Why? /


allocate int and double array

int *intP;double *doubleP;

// Allocate space for 10 integersintP = malloc(10 * sizeof(int));intP = malloc(10 sizeof(int));

// Allocate space for 10 doublesdoubleP = malloc(10 * sizeof(double));


allocate int and double array

int *intP;double *doubleP;

// Allocate space for 10 integersintP = malloc(10 * sizeof(int));

Allocates 40 bytessizeof(int) = 4

intP = malloc(10 sizeof(int));

// Allocate space for 10 doublesdoubleP = malloc(10 * sizeof(double));

Allocates 80 bytessizeof(double) = 8


realloc

realloc takes a pointer to allocated memory and reallocates the memory to a larger size

– if it can make the old block bigger, great– if not, it will get another, larger block, copy the old contents to the new contents free the old contents andcontents to the new contents, free the old contents and return a pointer to the new

intP = malloc(sizeof(int));intP = malloc(sizeof(int));intP = realloc(intP, 2*sizeof(intP));

intP may be different after a realloc!


int main () {double *dblPtr;int howMany randNum

Step 1: Prompt the user to enter the number of random numbers to generate

int howMany, randNum;

printf("How many random numbers to generate:");howMany=ProcessInput(stdin);

dblPtr = malloc(howMany * sizeof(double)); if (dblPtr == NULL){printf("memory allocation error exiting\n"); An example programprintf( memory allocation error, exiting\n );exit(1);}

p p g

for (int i=0;i<howMany;i++){randNum = random();dblPtr[i]=(randNum%10000)/1000 0;dblPtr[i]=(randNum%10000)/1000.0;}

PrintArray(dblPtr,howMany);


Step 2: create a dynamic array to store the random numbers

int main () {double *dblPtr;int howMany randNumint howMany, randNum;


dblPtr = malloc(howMany * sizeof(double)); if (dblPtr == NULL){printf("memory allocation error exiting\n");

In this example we have tested to be sure malloc

printf( memory allocation error, exiting\n );exit(1);}

succeeded. If not, we are out of memory.




Step 3: generate the random numbers and print them

int main () {double *dblPtr;int howMany randNumint howMany, randNum;


dblPtr = malloc(howMany * sizeof(double)); if (dblPtr == NULL){printf("memory allocation error exiting\n");printf( memory allocation error, exiting\n );exit(1);}




dblPtr = realloc(dblPtr, 2*howMany); Step 4: double the size of the array using realloc

for (int i=howMany; i<howMany*2; i++){{

randNum = random();dblPtr[i]=(randNum%10000)/1000.0;

}

Step 5: generate more random numbers and print them

howMany *= 2;

PrintArray(dblPtr howMany);PrintArray(dblPtr, howMany);free(dblPtr);

}


int ProcessInput(FILE *f)p ( ){int val; Safe data entry, just like last char in[100];fgets(in,100,f);

y, jweek

sscan(in, “%d”, &val);return val;

}


Print Array

void PrintArray(double *ptr, int cnt){printf("Printing Array Values\n");for(double *p=ptr; p<ptr+cnt; p++)( p p ; p p ; p )printf("Val is:%lf\n",*p);

}}


Documents

C-Michigan State University