CODEWITHSUNDEEP
c
Vbp
Vbp

Reputation: 1962

How to check address of pointer variable in C

I was trying to print the address of the pointer variable not the address where it is pointing to, could anyone assist me in achieving that? Below is what I am trying but it is showing warning which i am not able to resolve. Thanks!

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int* y;
    printf("%p\n",y);
    printf("%x\n",&y);
    y = (int*)malloc(sizeof(int));
    printf("%p\n",y);
    printf("%x\n",&y);
    return 0;
}

Compilation warning:

Warning: format ‘%x’ expects argument of type ‘unsigned int’,
    but argument 2 has type ‘int **’ 

Output:
0xb773fff4
bfa3594c
0x8361008
bfa3594c

Upvotes: 1

Views: 2155

Answers (2)

Bruno
Bruno

Reputation: 508

The code seems to compile without warning or error on Visual Studio 2012.

    #include <stdio.h>
    #include <stdlib.h>

    int _tmain(int argc, _TCHAR* argv[])
    {
        int* y = 0;
        printf("%p\n",y);
        printf("%x",&y);
        y = (int*)malloc(sizeof(int));
        printf("%p\n", y);
        printf("%x",  &y);

        return 0;
    }

The only recommendation is that you initialize y when you declare it.

Upvotes: -1

Jonathan Leffler
Jonathan Leffler

Reputation: 753715

Your second printf() should take a "%p\n" format, and strictly a cast:

printf("%p\n", (void *)&y);

The number of machines where the cast actually changes anything is rather limited.

Upvotes: 4

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