Pointer

POINTER

Consider the declaration,

Int i=3;

This declaration tell the C compiler to:

·        Reserve space in memory to hold the integer value.

·        Associate the name i with this memory location.

·        Store the value 3 at this location.

We can print this address number through following program:

#include<stdio.h>
int main()
{
    int i=3;
    printf("Address  of i = %u\n",&i);
    printf("Value of i = %d\n",i);
   printf(“Value of i = %d\n”,*(&i));
    return 0;
}

Output will:

Address of i = 2686748

Value of i = 3

Value of i = 3_

Hence it is printed out using %u which is a format specifier for printing an unsigned integer.

Note printing the value of *(&i) is same as printing the value of i.

The expression &i gives the address of the variable i. This address can be collected in a variable, by saying,

j=&i

But remember that j is not an ordinary variable like any other integer variable. It is a variable that contains the address of another variable (i in this case ).

i Contains the value 3 whose address is 2686748 and j’s value is i’s address i.e. 2686748.

Let see a program for better clearance…..

#include<stdio.h>
int main()
{
    int i=3;
    int *j;
    j=&i;
    printf("Address  of i = %u\n",&i);
    printf("Address  of i = %u\n",j);
    printf("Address  of j = %u\n",&j);
    printf("Value of j = %d\n",j);
    printf("Value of i = %d\n",i);
    printf("Value of i = %d\n",*(&i));
    printf("Value of i = %d\n",j);
    return 0;
}

Output 

Address of i = 2686748

Address of i = 2686748

Address of j = 2686744

Value of j = 2686748

Value of i = 3

Value of i = 3

Value of i = 3

_

Carefully observe the above program for better understanding.

Here is another program showing you how to use pointer in function call

#include<stdio.h>
int swap(int *,int *);
int main()
{
    int a=10,b=20;
    swap(&a,&b);
    printf("a = %d b = %d\n",a,b);
    return 0;
}
int swap(int *x,int *y)
{
    int t;
    t=*x;
    *x=*y;
    *y=t;
}

Output 

a =20 b = 10

_