How possible that an memory address also have another memory address ?

Question

I have stored the address of a integer variable in a pointer and then store the address of that previous into another pointer.I am not able to understand how it actually works.

#include <iostream>

using namespace std;

#include <stdio.h>

int main ()
{
    int  var;
    int  *ptr;
    int  **pptr;

    var = 30;

    /* take the address of var */
    ptr = &var;

   /* take the address of ptr using address of operator & */
    pptr = &ptr;

    /* take the value using pptr */
    printf("Value of var = %d\n", var );
    printf("Value available at ptr = %d\n", ptr );
    printf("Value available at pptr = %d\n", pptr);

    return 0;
}

possible duplicate of [Does a pointer also have any address or memory allocation?](http://stackoverflow.com/questions/991075/does-a-pointer-also-have-any-address-or-memory-allocation) — Santosh A, Apr 28 '15 at 10:58

score 5 · Accepted Answer · answered Apr 28 '15 at 10:59

When you do &ptr you get the address of the variable ptr is stored.

So you have a pointer pptr which points at ptr which in turn point at var. Like this:

+------+     +-----+     +-----+
| pptr | --> | ptr | --> | var |
+------+     +-----+     +-----+

On a side-note, don't use the "%d" format to print pointers. Use "%p" instead, and then cast the pointers to void *.

score 2 · Answer 2 · answered Apr 28 '15 at 11:05

Let's say var is stored at the memory address 0x12345678, ptr at the address 0x23456789 and pptr at the address 0x34567890.

When you do

var = 30;

you store 30 in the value of var (at the address of var, 0x12345678)

When you do

ptr = &var;

you store the address of var in the value of ptr. When you do

pptr = &ptr;

you store the address of ptr in the value of pptr.

The memory looks like this

Address       Value stored
...
0x12345678    30
...
0x23456789    0x12345678
...
0x34567890    0x23456789

If you try to print pptr, it will show 0x23456789 If you try to print *pptr, it will show the value at the address corresponding to the value of pptr, 012345678 And if you try to print **pptr, it will show the value at the address corresponding to the value at the address corresponding to the value of pptr, 30

haccks · Answer 3 · 2015-04-28T11:14:35.437

This ASCII art will help you to understand

       pptr                ptr                 var       

   +----------+        +----------+        +-----------+
   |          |        |          |        |           |
   | 0x67890  +------> |  0x12345 +------> |    30     |
   |          |        |          |        |           |
   +----------+        +----------+        +-----------+
     0xABCDE             0x67890              0x12345

How possible that an memory address also have another memory address ?

3 Answers3