ADDRESS ARITHMETIC:

ADDRESS ARITHMETIC:

1. The only possible operation on two pointer variables is subtraction. No other operation is possible. Whenever a subtraction operation is performed on two pointer variable, it gives the number of values that can be stored between those two addresses.

Eg.,

      int *p1,*p2;

Suppose p1=5000, p2=5040

Then, p2-p1=(5040-5000)= 40/2 =20.

i.e., 20 integer values can be stored between these two addresses.

Eg.,

float *p1,*p2;

Suppose p1=5000, p2=5040

Then, p2-p1=(5040-5000) = 40/4 =10.

i.e., 10 floating point values can be stored between these two addresses.

Eg.,

char *p1,*p2;

Suppose p1=5000,p2=5040

Then, p2-p1=(5040-5000)=40/1=40.

i.e., 40 character values can be stored between these two addresses.

2. Increment/decrement operations can be performed on pointer variables.

Eg.,

      int *p1;

Let p1=5000

p1++; or ++p1;

The incrementation operation causes the pointer variable increments to length of the datatype which is known as scaling factor.

      Hence, p1++ or ++p1 causes it to point to 5002. Since p1 is an integer pointer, an integer occupies 2 bytes, it increments by two locations.

Similarly the decrement operation causes the variable to point to the address after subtracting the scaling factor.

 Eg.,

      float *p1;

Let p1=5000

p1- -; or - -p1;

      After the decrement operation, p1 points to 4996.Since here the scaling factor is 4. Likewise, the scaling factor for character is 1.

3. When we add any integer value to the pointer variable, the integer vlue gets multiplied with the scaling factor and corresponding number of bytes will be added.

Eg.,

int *p1;

Suppose p1=5000

P1+10=5000+10

=5000+(10*2)

=5020

            Similarly, subtraction operation will be as follows:

Eg.,

float *p1;

Suppose p1=5000

P1-10=5000-10

   =5000-(10*4)

         =4960.

Ø  A Program for increment operation on pointers.

#include<stdio.h>

   #include<conio.h>

            void main( )

            {

int a=5,*p1;

p1=&a;

float b=15.5,*p2;

p2=&b;

char c=’a’,*p3;

p3=&c;

printf(“addresses are %u,%u,%u”,p1,p2,p3);

printf(“after incrementation\n”);

p1++;

p2++;

p3++;

printf(“addresses are %u,%u,%u”,p1,p2,p3);

getch( );

            }