Pointer to Pointer

Pointer is a variable that contains address of another variable. Now this variable itself might be another pointer. Thus, we now have a pointerthat contains another pointer’s address.

Example :

    #include <stdio.h>
    #include <conio.h>
    void main()
    {
    int i = 3, *j, **k ;
    j = &i ;
    k = &j ;

    printf ( "\nAddress of i = %u", &i ) ;
    printf ( "\nAddress of i = %u", j ) ;
    printf ( "\nAddress of i = %u", *k ) ;
    printf ( "\nAddress of j = %u", &j ) ;
    printf ( "\nAddress of j = %u", k ) ;
    printf ( "\nAddress of k = %u", &k ) ;
    printf ( "\nValue of j = %u", j ) ;
    printf ( "\nValue of k = %u", k ) ;
    printf ( "\nValue of i = %d", i ) ;
    printf ( "\nValue of i = %d", * ( &i ) ) ;
    printf ( "\nValue of i = %d", *j ) ;
    printf ( "\nValue of i = %d", **k ) ;
    }

Output :

Address of i = 65524
Address of i = 65524
Address of i = 65524
Address of j = 65522
Address of j = 65522
Address of k = 65520
Value of j = 65524
Value of k = 65522

Explanation :

Here the declarations as follows :

    int i = 3; //i is an ordinary int,
    int *j; //j is a pointer to an int
    int **k; //k is a pointer to an integer pointer

Now initializations as follow :

    j = &i ; //address of variable i is stored in j
    k = &j ;//address of variable j is stored in k

printf ("\nAddress of k = %u", &k); will return address of k.

printf ("\nAddress of j = %u", k); will return value of k or address of j.

printf ("\nAddress of i = %u", *k); will return the address of i.

*k = *(&j)=>*(65522) will return value stored at this address of the variable j i.e 65524 = 65524

printf ("\nValue of i = %d", **k); will return the value of i.

**k = **(&j)=>[*(&j)] will give the value stored at the address of variable j i.e 65524 which is the address of variable i i.e &i.

= *(&i) will give value stored at the address of variable i i.e 3.