How the address of pointer that holds the address of array are same?

Question

Here p is a integer pointer that can hold the address of int variable, but it also has a memory address - where it is stored.

let base address of array a = 1002 address of pointer p = 2008

when we write: int *p=a; //p points to the base address of array a int *p=a; //p points to the base address of array a
and int **r=&p; //means *r points to the address of p int **r=&p; //means *r points to the address of p

how *r points to the address of a , it should point to address of p .

#include <stdio.h>
void main()
{
    int a[3] = {1, 2, 3};
    int *p =a;
    int **r = &p;
    printf("%p %p", *r, a);
}

Answer 1

Your printf in not correct. It should be r to print the address r points to:

printf("%p %p", r, a);

By using *r , you deference r (ie, jump to the address r is pointing to) and thus printing the address of a .

Answer 2

Please note that

int *p=a;

means that a and p are now pointing to same address.

Also, it is r that points to p (is the address of p ) and not *r .

Thus to print the address of p simply use r instead of *r in printf() .

printf("%p %p", r, a);

Answer 3

What is below?

 int **r = &p;

Basically using above, r holds address of p right? So if you dereference r , like you do *r , it will try to retrieve value stored at address of p right? Which is a .

Note: You need to cast to void* in printf:

 printf("%p %p", (void*)*r, (void*) a);

Answer 4

Isn't ->-> same as ->

When you say,

int *p = a;

it means, P is pointing to a, or P holds the address of a.

Same in case of int **r=&p;

R holds the address of P, had you used printf("%p %p", r, a); , you would have got the address of P.

but since you are dereferencing r, you got the address of a.