Lecture 10 Introduction to Programming in C Prof. Dr. Arne Kutzner Hanyang University / Seoul Korea

Lecture 10

Introduction to Programming in C

Prof. Dr. Arne Kutzner

Hanyang University / Seoul Korea

Pointers / Introduction

Introduction to C L10.3

Variables and Memory Address

• Every variable has some location in the memory that holds the value of the variable. This location is called the address of the variable.

1000

0 1004

1008

1012

1016

1020

int num = 0;

num

The address of the variable num

The address of the variable num

Introduction to C L10.4

Pointers

• A pointer is also a variable, but unlike the other types of variables that we have seen, it contains a memory address as its value.

Introduction to C L10.5

Pointer Declaration

• To declare a pointer variable, a asterisk(*) should placed in front of name of the name of the variable. Example:

int *p;

p is a pointer to an integer value (i.e. p contains the address of some memory cell that contains an integer value)

Introduction to C L10.6

Address Operator

• You can get the memory address of some variable by using the address operator &. Example:int x = 0;int *p;p = &x;

The pointer p gets the address of the variable x as its value

1000

0 1004

1008

1012

1004 1016

1020

x

p

Introduction to C L10.7

How to deference pointers?

• For dereferencing a pointer place a asterisk in front of the pointer variable name. Example:int x = 0;int *p;p = &x;*p = 5;

Write the value 5 to the memory location referenced by p

1000

5 1004

1008

1012

1004 1016

1020

x

p

Introduction to C L10.8

Pointers / More explanation

• So we say:p is a pointer to the integer variable x.

x 0

p 1004

*p num 0

&x 1004

The * operator is referred to as indirection or

dereferencing operator

The * operator is referred to as indirection or

dereferencing operator

p

0

x

*p is equal to 0 (the value of x).p indirectly references the value 0.

Introduction to C L10.9

Pointer Types

• Pointers can be declared to point to variables of any type.

• Examples:int *p1; pointer to an integerdouble *p2; pointer to a floating point numberchar *p3; pointer to a character

Introduction to C L10.10

Example Program 1int main() {…int x1 = 5;

int x2 = 8;

int *x1ptr;

int *x2ptr;

x1ptr = &x1;

x2ptr = &x2;

*x2ptr = 12;…}

x1 5 1000

x2 8 1004

x1ptr ? 1008

x2ptr ? 1012

x1 5 1000

x2 8 1004

x1ptr 1000 1008

x2ptr 1004 1012

x1 5 1000

x2 12 1004

x1ptr 1000 1008

x2ptr 1004 1012

Introduction to C L10.11

Example Program 2

1 #include <stdio.h>2 int main() {3 int num1, num2, product;4 int *ptr_num1, *ptr_num2;

5 num1 = 5;6 num2 = 7;

7 ptr_num1 = &num1;8 ptr_num2 = &num2;

9 product = num1 * (*ptr_num2); … }

num1 5 1000

num2 7 1004

product 35 1008

ptr_num1 1000 1012

ptr_num2 1004 1014

The resulting memory picture

* operator specifies multiplication

* operator specifies multiplication

* operator specifies dereferencing a pointer

* operator specifies dereferencing a pointer

Introduction to C L10.12

Description of Example Program 2

• Line 3: integer variable declaration• Line 4: pointer to integer declaration• Line 7: The address of num1 is

evaluated by the & operator and stored in ptr_num1

• Line 9: The value in num1 is multiplied by the value of the address location pointed by ptr_num2

Introduction to C L10.13

Pointer type equivalence

• A pointer can dereference only a variable which is of the same type.Example:int n = 5;double x = 5.0;int *ptr1;double *ptr2;ptr1 = &n;ptr2 = &n;ptr1 = &x;

Compile error!ptr1 is a pointer to integer whereas x is a floating point number.ptr2 is a pointer to double whereas n is an integer number.

Compile error!ptr1 is a pointer to integer whereas x is a floating point number.ptr2 is a pointer to double whereas n is an integer number.

Introduction to C L10.14

Warning:Uninitialized Pointers

• Dereferencing a pointer that has not been properly initializedExample:

int *p; *p = 1;printf("Value is %d", *p);

The pointer contains an arbitrary address, so *p=1 may overwrite some arbitrary location in memory which can cause severe problems.Run-time error!No compile error!

Introduction to C L10.15

Effects with uninitialized pointers

• Program attempts to access some random memory location.Possible effects:– Fatal Runtime error (common program run-

time error message: "segmentation fault").– Accidentally modifying other data /

variables. Program continues but delivers incorrect result.

Functions and Pointers

Introduction to C L10.17

Pointers and Functions

• Function arguments can be of some “pointer type”. Example:void main () { int i;

fun (&i); printf("%d", i) } void fun (int *p) { *p = 1;}

Type of function arguments is “pointer to int”

Introduction to C L10.18

Call by Reference

• In the example on the slide before the statement *p = 1 in fun changes the value of i defined main in.

• Because p refers i in fun we call this form of argument passing call by reference.

Introduction to C L10.19

Example for the Application of Call by Reference

• A function fun shall have two input arguments and two output argument.

(Remember: Using return we can deliver only one value!)

fun

x

y

x + y

x * yInput Output

Introduction to C L10.20

Application of Call by Reference

int main() { int x, y, sum, product; x = 3; y = 5; fun (x, y, &sum, &product)}

void fun (int a, int b, int *p1, int *p2){ *p1 = a + b; *p2 = a * b;}

The addresses of sum and product are passed as arguments.

Introduction to C L10.21

Application of Call by Reference

• In the previous example, the addresses (not values) of the variables sum and product are copied to the function pointer arguments p1 and p2. In this way, sum and product have the same memory locations as *p1 and *p2, respectively.

x 3

y 5

p1 1476

product 15 1468

p2 1468

sum 8 1476

….

a 3

b 5

Introduction to C L10.22

Comparison of Call by Reference and Call by Value

• With the function call by value, the values of the function arguments can't be modified by the function.

• With the function call by reference, the function arguments are the addresses, not the values of the corresponding variables. These addresses are copied into some pointers which are function arguments. Therefore, modifying the values stored in the addresses pointed by these pointers modifies also the values of the variables.

Introduction to C L10.23

Comparison of Call by Reference and Call by Value

Call by value…int n = 1;int m = 2;

swap(n, m);printf("%d, %d", n, m);…

void swap(int x, int y) int temp = x; x = y; y = temp;}

Call by reference…int n = 1;int m = 2;

swap(&n, &m);printf("%d, %d", n, m);…

void swap(int *x, int *y) int temp = *x; *x = *y; *y = temp;}

What outputs do we get and why?

Introduction to C L10.24

Remark: scanf

• Now you should be able to understand the scanf function:

scanf("%d", &num);

Memory address of num

Pointers and Arrays /Pointer Arithmetic

Introduction to C L10.26

Pointers and Arrays

• In C there is a strong relationship between the concepts of pointers and arrays

• An array name is basically a const pointer. (A pointer with fixed address)int *x;

int a[10];

x = a;

This statement is OK!!!It assigns x the address of the first elements array a.

Introduction to C L10.27

Pointers and Arrays (cont.)

• You can even use the [ ] operator with a pointer:int *x;

int a[10];

x = &a[2];

for (int i = 0;i < 3 ;i++)

x[i]++;

x gets “the address of a[2] ”

x[i] is identical to a[i+2]

Introduction to C L10.28

Pointer arithmetic• Integer math operations can be used with

pointers.

• If you increment a pointer, it will be increased by the size of whatever it points to.

int a[5];

a[0] a[1] a[2] a[3] a[4]

int *ptr = a;

*ptr

*(ptr+2)*(ptr+4)

Introduction to C L10.29

printing an array

void print_array(int a[], int len) {

for (int i = 0;i < len; i++)

printf ("[%d] = %d", i, a[i]);

}

void print_array(int *a, int len) {

for (int i = 0;i < len; i++)

printf ("[%d] = %d", i, *a++);

}

pointer version

array version

Introduction to C L10.30

Arrays of Pointers

• Like for any other type you can create arrays of pointers:

int a[] = {0, 1, 2};int b[] = {4, 5, 6};int* x[2];x[0] = a;x[1] = b;

printf("%d",*(x[1]+1));printf("%d",*(*(x+1)+1));

x is array of pointers

both statementsare identical

Introduction to C L10.31

Matrices and Pointers …• The technique can be extended to

multidimensional arrays:int a[2][3] = {{0, 1, 2},

{4, 5, 6}};int* x[2];x[0] = &a[0];x[1] = &a[1];

printf("%d",*(x[1]+1));printf("%d",*(*(x+1)+1));printf("%d",a[1][1]);printf("%d",*(a[1]+1));printf("%d",*(*(a+1)+1));

all statementsdeliver the same element

Introduction to C L10.32

Pointers to Pointers

int *x;

int **y;

x some int

some *int some inty

• Pointer can refer pointers …

Strings

Introduction to C L10.34

Strings

• Recall that C data type char is used to represent digits, letters, and characters

char letter = 'A';• A string is a sequence of characters. In C we

use arrays of type char for the representation of strings. Example:

char color[] = "blue";• The above statement declares a character

array, that is a string, and initializes it to "blue".

Introduction to C L10.35

Strings

• Let's see what happens to memory after this declaration and initialization

b l u e \0 The null characterThe null character

• Note that the string contains the character '\0'; the null character that marks the end of the string.

• The same string could be also initialized as follows:char color[]={'b','l','u','e','\0'};

Introduction to C L10.36

Strings

• Another possibility is to fix the size of the string in declaration:

char color[10]="blue";

b l u e \0 ? ? ? ? ?

• Make sure that the size of the character array is larger than the number of characters in the string plus the null character.

char color[4] = "blue";char color[5] = "blue"; No space left for

the null character

No space left for the null character

OKOK

Introduction to C L10.37

String Initialization

char str1[10] = "green";

char str2[10] = {'g','r','e','e','n','\0'};

char str3[10];str3 = "green"; /* !! ERROR !! */

OKOK

• You can initialize a string only in the context of its declaration.

Introduction to C L10.38

Strings with printf and scanf• The scanf function, when it scans a string,

skips leading whitespace characters such as blanks, new lines and tabs. When it comes across another whitespace character, it stops and places the null character at the end of the string.

• If one tries to print a string having no null character at the end with printf, printf continues to display the contents of memory locations until it encounters a null character or until it attempts across a forbidden memory cell.

Introduction to C L10.39

String Input/Output withprintf & scanf

#include <stdio.h>#define STRING_LEN 10int add (int a, int b);int main() { char dept[STRING_LEN], days[STRING_LEN]; int course_num, time; printf("Enter department code, course number, days "); printf("time like this: EB 1003 Thur 1000\n>"); scanf("%s%d%s%d", dept, &course_num, days, &time); printf("%s %d is held %s at %d\n", dept, course_num, days, time);

return 0;}

Introduction to C L10.40

String Library Function: strcpy• The C library provides many functions to manipulate

strings; #include string.h• strcpy:

Prototype:char *strcpy (char *dest, char *source);Purpose: Makes a copy of source, a string, in the character array accessed by dest.

• Example:char str1[10];strcpy(str1, "EB1003-01")

E B 1 0 0 3 - 0 1 \0str1:

Introduction to C L10.41

String Library Function: strncpy• strncpy:

Prototype:char *strncpy (char *dest, char *source, unsigned n);Purpose: Makes a copy of up to n characters from source in dest.

• Example: char str1[10], str2[20]="SKU EB1003-01"; strncpy(str1, str2, 9); str1[9] = '\0';

S K U E B 1 0 0 \0str1:

Introduction to C L10.42

String Library Function: strncpy• strncpy can be used to extract a number of

characters from a string. For example, you might want to examine course code in the string "EB1003-01"

• Example: char str1[10] = "EB1003-01"; char course_code[5]; strncpy(course_code, &str1[2], 4); course_code[4]='\0';

E B 1 0 0 3 - 0 1 \0str1:

1 0 0 3 \0course_code:

Introduction to C L10.43

String Library Function: strcat• strcat:

Prototype:char *strcat (char *dest, char *source);Purpose: Appends source to the end of dest.

• Example: char str1[10] = "EB";

E B \0 ? ? ? ? ? ? ?str1:

strcat(str1, "1003-01");

E B 1 0 0 3 - 0 1 /0str1:

Introduction to C L10.44

String Comparison• Standard C library also provides strcmp

function to compare strings.• strcmp:

Prototype:int strcmp(char *s1, char *s2);Purpose: Compares s1 and s2 alphabetically, returns a negative value if s1 should precede s2, zero if s1 and s2 are equal, and a positive value if s2 should precede s1 in an alphabetized list.Example: char s1[10] = "Pusan", s2[10] = "Seoul"; if (strcmp(s1, s2) < 0)

Introduction to C L10.45

String Comparisons

strcmp ("thrill", "throw"); returns a negative integer

strcmp ("read", "readable); returns a negative integer

strcmp ("shrimp", "crab"); returns a positive integer

strcmp ("end", "end) returns zero

Introduction to C L10.46

String Comparison• strncmp:

Prototype:int strncmp(char *s1, char *s2, unsigned n);Purpose: Compares the first n characters of s1 and s2 returning positive, zero, and negative values as does strcmp.Example:

strncmp ("read", "readable", 1);strncmp ("read", "readable", 2);strncmp ("read", "readable", 3);strncmp ("read", "readable", 4);

All returns zero

Introduction to C L10.47

Scanning a Full Line• With scanf, the strings containing spaces can

not be input at once.• C provides the gets function to read a line of

text into a string.

char string1[20];gets(string1);

• The gets function waits for the user to input some text and finalized by the enter key. For example, if we type "text string" and press enter, string1 will be set to "text string" at once.

Introduction to C L10.48

Arrays of Strings

• Since a string is an array of characters, an array of strings is a two dimensional array of characters where each row is a string.

char week[7][10] = {"Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"};

printf("Today is %s.", week[0]);

Monday

Introduction to C L10.49

Character Pointers

char a[] = "Hello";

H e l l o \0a:

char *p = "Hello";

H e l l o \0p:

Introduction to C L10.50

Arrays of Pointers to charschar s1[]="March"; char s2[]="April";char s3[]="May";

char *string[3];

string[0] = s1;string[1] = s2;string[2] = s3;

spring

March\0

April\0

May\0

Introduction to C L10.51

Character Analysis

• In many string-processing applications, we need to know if a character is a letter, a digit or a punctuation mark. The library "ctype.h" defines such facilities:

• isalpha: If argument is a letter of the alphabet.• isdigit: if argument is one of the ten decimal digits.• islower (isupper): if argument is a lowercase (or

uppercase) letter of the alphabet.• ispunct: if argument is a punctuation character.• ispace: if argument is a whitespace character such

as a space, newline, or a tab.

The Heap /Dynamic Memory Allocation

Introduction to C L10.53

sizeof(...) function

• Using the sizeof function we can get the number of bytes consumed by instances of some datatypeSyntax: sizeof(<datatype name>);Example:printf("%d", sizeof(int));

Introduction to C L10.54

The Heap

• The Heap is a special area of memory that allows the dynamic allocation of memory during Runtime.

• Possible Application:Arrays with dynamic size, e.g. arrays where we know the size only during runtime(Remember: So far we have to specify the size (number of elements) of some array in the context of its declaration. E.g.: x[10])

Introduction to C L10.55

Heap Management in C

• Three functions for heap management defined in stdlib.h– calloc(n, el_size) – allocation of continuous

space for an array of n elements with el_size bytes for each array element

– malloc(size) – allocation of an area of size bytes in the heap

– free(p) – freeing of allocated heap-space. (C stores internally the amount of memory associated with p)

Introduction to C L10.56

Warning: Dangling Pointer Problem

• Don’t use pointers that refer to some freed memory area

• Example:

int *p;p = malloc(sizeof(int));…free(p);… *p = 5;

p is here a “dangling pointer”. This statement can create serious trouble.

Introduction to C L10.57

“void” pointers in C

• malloc returns a pointer of type void*. The type-checker of C is quite tolerant with this type of pointers. Example:The following code generates no error during compile time (not even a warning)!

void* fun () {return NULL;

}void main() {

int *p;p = fun();

}