How to check address of pointer variable in C

Question

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

You took its address with & correctly -- the printf hints are just hints — Warren P
– Warren P, Commented Dec 6, 2012 at 3:48
Both y and &y are pointer values. Why are you using %p to print one pointer value, but suddenly switch to %x for another pointer value? All of your printf statements are printing pointers, which means that all of them should use %p. — AnT stands with Russia
– AnT stands with Russia, Commented Dec 6, 2012 at 3:59
Also, I find it hard to believe that the output you posted was produced by the code you posted. You posted either wrong code, or wrong output. — AnT stands with Russia
– AnT stands with Russia, Commented Dec 6, 2012 at 4:03
@AndreyT Sorry, you got it right I have to edit the output, I was trying too much with %x and %p to print the address of the pointer variable itself, and somewhere in between posted the code in hurry I am correcting it. Appreciate your assistance! — Vbp
– Vbp, Commented Dec 6, 2012 at 4:14

Jonathan Leffler · Accepted Answer · 2012-12-06 03:50:23Z

4

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.

answered Dec 6, 2012 at 3:50

Jonathan Leffler

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2 Comments

julumme Over a year ago

As a comment, %x is unsigned int, and for example on 64bit machines, this will not be same as %p. As Jonathan said, always %p for pointers

Jonathan Leffler Over a year ago

I'm glad AndreyT asked you about the output because the original output didn't make much sense. Unfortunately, I was helping my kids with homework at the same time and didn't get to raise the question. Be aware that the format for %p varies between platforms. If you want to control the pointer format (I usually do), then use <stdint.h> and printf("0x%.8" PRIXPTR "\n", (muintptr_t)&y);. This uses string concatenation and the macros defined in the header to format the address with at least 8 digits and with upper-case hex digits and the 0x prefix. Not all platforms add the 0x prefix.

Bruno · Accepted Answer · 2012-12-06 03:55:54Z

-1

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.

answered Dec 6, 2012 at 3:55

Bruno

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1 Comment

dave Over a year ago

I think your answer is a sign that Visual Studio is broken, not that the code is fine. Perhaps you need to turn on more warnings - the warning is valid!

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