Instructor: Alexander Stoytchev alex/classes/2009_Fall_185/ CprE 185: Intro to Problem Solving (using C)

Instructor: Alexander Stoytchev

http://www.ece.iastate.edu/~alex/classes/2009_Fall_185/

CprE 185: Intro to Problem Solving

(using C)

Formatting Numbers

CprE 185: Intro to Problem SolvingIowa State University, Ames, IACopyright © Alexander Stoytchev

Administrative Stuff

• HW3 is out

• Due on Friday Sep 18 @ 8pm Electronic submission on WebCT

• HW2 is due this Friday @ 8pm

Quick review of the last lecture

On-line IEEE 754 Converters

• http://www.h-schmidt.net/FloatApplet/IEEE754.html

• http://babbage.cs.qc.edu/IEEE-754/Decimal.html

http://www.h-schmidt.net/FloatApplet/IEEE754.html

http://babbage.cs.qc.edu/IEEE-754/Decimal.html

Expressions

• An expression is a combination of one or more operators and operands

• Arithmetic expressions compute numeric results and make use of the arithmetic operators:

• If either or both operands used by an arithmetic operator are floating point, then the result is a floating point

AdditionSubtractionMultiplicationDivisionRemainder

+-*/%

© 2004 Pearson Addison-Wesley. All rights reserved

Division and Remainder

• If both operands to the division operator (/) are integers, the result is an integer (the fractional part is discarded)

• The remainder operator (%) returns the remainder after dividing the second operand into the first

14 / 3 equals

8 / 12 equals

4

0

14 % 3 equals

8 % 12 equals

2

8


Operator Precedence

• Operators can be combined into complex expressions

result = total + count / max - offset;

• Operators have a well-defined precedence which determines the order in which they are evaluated

• Multiplication, division, and remainder are evaluated prior to addition, subtraction, and string concatenation

• Arithmetic operators with the same precedence are evaluated from left to right, but parentheses can be used to force the evaluation order


Operator Precedence

• What is the order of evaluation in the following expressions?

a + b + c + d + e1 432

a + b * c - d / e3 241

a / (b + c) - d % e2 341

a / (b * (c + (d - e)))4 123


Expression Trees

• The evaluation of a particular expression can be shown using an expression tree

• The operators lower in the tree have higher precedence for that expression

a + (b – c) / da

+

/

- d

b c


Assignment Revisited

• The assignment operator has a lower precedence than the arithmetic operators

First the expression on the right handside of the = operator is evaluated

Then the result is stored in thevariable on the left hand side

answer = sum / 4 + MAX * lowest;

14 3 2


Effect of sum = sum + item;

Figure 2.4. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Effect of scanf("%lf", &miles);


Evaluation Tree for area = PI * radius * radius;


Step-by-Step Expression Evaluation


Evaluation Tree and Evaluation for v = (p2 - p1) / (t2 - t1);


Evaluation Tree and Evaluation for z - (a + b / 2) + w * -y


Assignment Revisited

• The right and left hand sides of an assignment statement can contain the same variable

First, one is added to theoriginal value of count

Then the result is stored back into count(overwriting the original value)

count = count + 1;


Increment and Decrement

• The increment and decrement operators use only one operand

• The increment operator (++) adds one to its operand

• The decrement operator (--) subtracts one from its operand

• The statement

count++;

is functionally equivalent to

count = count + 1;


Increment and Decrement

• The increment and decrement operators can be applied in postfix form:

count++

• or prefix form:

++count

• When used as part of a larger expression, the two forms can have different effects

• Because of their subtleties, the increment and decrement operators should be used with care


Assignment Operators

• Often we perform an operation on a variable, and then store the result back into that variable

• C provides assignment operators to simplify that process

• For example, the statement

num += count;

is equivalent to

num = num + count;



• There are many assignment operators in C, including the following:

Operator

+=-=*=/=%=

Example

x += yx -= yx *= yx /= yx %= y

Equivalent To

x = x + yx = x - yx = x * yx = x / yx = x % y



• The right hand side of an assignment operator can be a complex expression

• The entire right-hand expression is evaluated first, then the result is combined with the original variable

• Therefore

result /= (total-MIN) % num;

is equivalent to

result = result / ((total-MIN) % num);


Escape Sequences

• Some C escape sequences:

Escape Sequence

\b\t\n\r\a\"\'\\

Meaning

backspacetabnewlinecarriage returnbeepdouble quotesingle quotebackslash


printf() function

• printf(“format string”, variable1, variable2, …);

• printf(“For int use %d”, myInteger);

• printff(“For float use %f”, myFloat);

• printf(“For double use %lf”, myDouble);

• printf(“For float or double %g”, myF_or_D);

• printf(“int=%d double %lf”, myInteger, myDouble);

scanf() function

• scanf(“format string”, &variable1, &variable2, …);

• scanf(“%d”, &myInteger);

• scanf(“%f”, &myFloat);

• scanf(“%lf”, &myDouble);

• scanf(“%d%f”, &myInteger, &myFloat);

Common Bugs

• Using & in a printf function call.

printf(“For int use %d”, &myInteger); // wrong

• Using the wrong string in printf

printf(“This is a float %d”, myFloat); // use %f not %d

• Not using & in a scanf() function call.

scanf(“%d”, myInteger); // Wrong

• Using the wrong string in scanf()

scanf(“%d”, &myFloat); // wrong; use %f instead of %d

Why do we need to specify the format of the keyboard input before we can read it?

printf() arguments

• %[flags][width][.precision][length]specifier

• http://www.cplusplus.com/reference/clibrary/cstdio/printf.html

Printing octal, hexadecimal and binary

Memory Analogy

Address 0

Address 1

Address 2

Address 3

Address 4

Address 5

Address 6

Memory Analogy (32 bit architecture)

Address 0

Address 4

Address 8

Address 12

Address 16

Address 20

Address 24

Memory Analogy (32 bit architecture)

Address 0x00

Address 0x04

Address 0x08

Address 0x0C

Address 0x10

Address 0x14

Address 0x18

Hexadecimal

Address 0x0A

Address 0x0D

Storing a Double

Address 0x08

Address 0x0C

Storing 3.14

• 3.14 in binary IEEE-754 double precision (64 bits)

sign exponent mantissa

0 10000000000 1001000111101011100001010001111010111000010100011111

• In hexadecimal this is (hint: groups of four):

0100 0000 0000 1001 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1111

4 0 0 9 1 E B 8 5 1 E B 8 5 1 F

Storing 3.14

• So 3.14 in hexadecimal IEEE-754 is 40091EB851EB851F

• This is 64 bits.

• On a 32 bit architecture there are 2 ways to store this

Small address: 40091EB8 51EB851F

Large address: 51EB851F 40091EB8

Big-Endian Little-Endian

Storing 3.14

Address 0x08

Address 0x0C

40 09 1E B8

51 EB 85 1F

Address 0x08

Address 0x0C 40 09 1E B8

51 EB 85 1F

Big

-End

ian

Litt

le-E

ndia

n

Storing 3.14 on a Little-Endian Machine(these are the actual bits that are stored)

Address 0x08

Address 0x0C

01010001 11101011 10000101 00011111

01000000 00001001 00011110 10111000

Once again, 3.14 in IEEE-754 double precision is:

sign exponent mantissa

0 10000000000 1001000111101011100001010001111010111000010100011111

They are stored in binarythe hexadecimals are just for visualization

Address 0x08

Address 0x0C4 0 0 9 1 E B 8

5 1 E B 8 5 1 F01010001 11101011 10000101 00011111

01000000 00001001 00011110 10111000

Big-Endian

(8-bit architecture)


http://en.wikipedia.org/wiki/Endianness

Little Endian

http://en.wikipedia.org/wiki/Endianness



Big-Endian/Little-Endian analogy

[image fom http://www.simplylockers.co.uk/images/PLowLocker.gif]





What would be printed?(don’t try this at home)

double pi = 3.14;

printf(“%d”,pi);

• Result: 1374389535

Why? 3.14 = 40091EB851EB851F (in double format) Stored on a little-endian 32-bit architecture

• 51EB851F (1374389535 in decimal)• 40091EB8 (1074339512 in decimal)


double pi = 3.14;printf(“%d %d”, pi);

• Result: 1374389535 1074339512



• The second %d uses the extra bytes of pi that were not printed by the first %d


double a = 2.0;

printf(“%d”,a);

• Result: 0

Why? 2.0 = 40000000 00000000 (in hex IEEE double format) Stored on a little-endian 32-bit architecture

• 00000000 (0 in decimal)• 40000000 (1073741824 in decimal)

What would be printed?(an even more advanced example)

int a[2]; // defines an int arraya[0]=0;a[1]=0;scanf(“%lf”, &a[0]); // read 64 bits into 32 bits// The user enters 3.14printf(“%d %d”, a[0], a[1]);

• Result: 1374389535 1074339512



The double 3.14 requires 64 bits which are stored in the two consecutive 32-bit integers named a[0] and a[1]

Questions?

THE END

Documents

Instructor: Alexander Stoytchev alex/classes/2009_Fall_185/ CprE 185: Intro to Problem Solving (using C)