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
and 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);
}

Why wouldn't it be the same ? You let r point to the address where p is stored and then print *r, so obviously that is p, which you assigned to be a — marcolz
– marcolz, Commented Jul 28, 2015 at 11:56
It should be int *p = &a; or else it is an invalid conversion — Devang Jayachandran
– Devang Jayachandran, Commented Jul 28, 2015 at 11:59

Matthias J. Sax · Accepted Answer · 2015-07-28 11:56:23Z

7

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.

answered Jul 28, 2015 at 11:56

Matthias J. Sax

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TryinHard · Accepted Answer · 2015-07-29 11:13:41Z

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);

edited Jul 29, 2015 at 11:13

answered Jul 28, 2015 at 11:56

TryinHard

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gmoniava · Accepted Answer · 2015-07-28 12:03:13Z

2

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);

answered Jul 28, 2015 at 12:03

gmoniava

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Nishant · Accepted Answer · 2015-07-28 12:06:14Z

1

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.

answered Jul 28, 2015 at 12:06

Nishant

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