Programming and Data Structures

306 Programming and Data Structures

OUTPUT:

llll

2 2 2 2

3 3 3 3

Explanation In the above example, addresses of one-array elements are assigned to another array.

The second array assigns addresses to the third one and so on. In the successive printf () statements

increases by one for getting higher level value.

9.9 POINTERS AND STRINGS

9.29 Write a program to read string from keyboard and display it using character pointer.

#Include <stdio.h>

#Include <conio.h>

nainO

{

char name [15] ,*ch;

printf ("Enter Your Name

getsiname);

ch=mame;

J* store base address of string name */

while (*ch !-'\0 ')

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printf C % c ” *ch);

ch++;

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OUTPUT:

Enter Your Name: KUMAR

KUMAR

Explanation Here, the address of 0 th element is assigned to the character pointer ' ch'. In other

words base address of string is assigned to ' ch' . The pointer ' * ch' points to the value stored at that

memory location and it is printed through p rin tf () statement. After every increase in 'c h ' the

pointer goes to the next character of the string. When it encounters NULL character the w hile loop

terminates the program.

9.30 Write a program to find length of a given string including and excluding spaces using

pointers.

# include <stdio.h>

# include <ccnio.h>

void main ()

{

char str [20] ,*s?

int p»0,q*0;

Pointers 307

d ta c x O ;

printf ("Enter String

gets(str);

s=str;

while (*s!='\0r)

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printf ("% c "* s );

p++;

S++/

if(*s==32) /* ASCII equivalnet o f " (space)is 32*t

q++!

printf ("\nLength of Siring including spaces: %d",p);

printf ("\nhength of String excluding spaces : %d",p-q);

I

OUTPUT:

Enter String: POINTERS ARE EASY

POINTERS ARE EASY

Length of String including spaces: 17

Length of String excluding spaces: 15

Explanation The above program is the same as the previous one. Here, the counter variables ' p '

and ' q' are increased to count number of characters and spaces found in the string. The ASCII value

of space is 3 2. Thus, at the end of the program both the variables are printed.

931 Write a program to interchange elements of character array using pointer.

# include <stdio.h>

# include <conio.h>

void main ()

{

char *xxames [] *{

"kapil*,

"maaoj",

"amit",

"amol",

"pavan",

"mahesh"

} ;

char *tmp;

clrscrO ;

printf ("Original: %s %s",names[3],names[4]);

tmp=names[3];

namesl3]=names[4];

names[4]=tmp;

printf ("\nNew : %s %s",names[3],names[4]);

}

OUTPUT:

Original: amol pavan

New : pavan amol

308 Programming and Data Structures

Explanation In the above program character array *names [ ] is declared and initialized. Another

character pointer *tmp is declared. The destination name that is to be replaced is assigned to the

variable ' *tm p '. The destination name is replaced with the source name and the source name is

replaced with * trap variable. Thus, using simple assignment statements two names are interchanged.

932 Write a program to read two strings through the keyboard. Compare these two strings

Character by character. Display the similar characters found in both the strings and

count the number of dissimilar characters.

# include <stdio.h>

# include <canio.h>

# include <string.h>

void main ()

{

char strl[20] ,str2[20] ,l,*a ,* b ;

Int cxO;

clrscr ();

printf ( “\ti Enter First String :");

gets(strl);

printf ("\n Enter Second String

gets(strl);

a=strl;

b=str2;

printf ( “\n Similar Characters Found in Both String.");

while (*a!='\0')

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if(stricmp(*a,*b)=±0)

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printf ("\n\t%c \t%c”,*a,*b);

/++;

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else

C + + ;

a++;

b++;

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if (c==0)

printf ("\n The String are Identical.’’);

else

printf ("\nThe Strings are different at %d places",c);

printf ("\n The String Characters are similar at % d places.",!);

OUTPUT;

Enter First String : SUNDAY

Enter Second String: M O NDAY

Similar Characters Found in Both String.

N N

D D

A A

Y Y

The Strings are different at 2 places.

The String Characters are similar at 4 places.