print the memory location of a variable (or pointer)

Question

I want to print where a variable is stored. I Google it and I found this:

int *p;
printf("memory location of ptr: %p\n",  (void *)p);

If I write this, is it right?

printf("memory location of ptr: %p\n",  &p);

I compiled it and I didn't get any errors or warnings. However, the above two commands didn’t return the same value!

Answer 1

Lets say you have these declarations:

int i;
int *p = &i;

It would look something like this in memory:

+---+     +---+
| p | --> | i |
+---+     +---+

If you then use &p you get a pointer to p, so you have this:

+----+     +---+     +---+
| &p | --> | p | --> | i |
+----+     +---+     +---+

So the value of p is the address of i, and the value of &p is the address of p. That's why you get different values.

Answer 2

if I write this, is it right?

No. This is not right. For %p specifier you must have to cast it to void *. Your compiler should give warning about this:

warning: format ‘%p’ expects argument of type ‘void *’, but argument 2 has type ‘int **’ [-Werror=format=]  

Read this answer. It says that:

The %p format requires an argument of type void*. If pointers of type int* and int(*)[10] have the same representation as void* and are passed as arguments in the same way, as is the case for most implementations, it's likely to work, but it's not guaranteed. You should explicitly convert the pointers to void*

Draft n1570; 7.21.6 Formatted input/output functions:

p The argument shall be a pointer to void. The value of the pointer is converted to a sequence of printing characters, in an implementation-defined manner.

---

but theses two statements don’t return the same value if I put both of them in the program.

Yes,it will not return the same value. P will give you the value of address of the variable p points to (pointee) while &p will give you the value of the address of pointer p itself. Cast &p to void *.

printf("memory location of ptr: %p\n",  (void *)&p);  

Answer 3

If you use the "&" operator on a pointer, you create a pointer to the pointer. So the result in your case would be an int**.

printf expects a void* when you use the %p specifier, therefore the first way is correct.

Answer 4

As you have discovered, the two don't do the same thing.

The first will print the value stored in p. You would usually expect this to be the address of an int (though people do the darnedest things).

The second will print the address of p. However, you need to add a cast to void * for portability reasons, so the second should read

printf("memory location of ptr: %p\n", (void *)&p);

I want to print where a variable is stored.

So the answer to your question depends on which variable you are concerned with? For the int that p points to you want to use the first one. For the variable p itself you want to use the second.

Answer 5

#include <stdio.h>

int main()
{
    int i;
    int *x = &i;

    i = 8;
    printf("value of i is %d\n",i);
    printf("value of x is %d\n", *x);
    printf("address of x is %p\n",x);
    printf("address of i is %p\n",&i);
}

output:

~/workspace/C$ ./test

value of i is 8

value of x is 8

address of x is 0x7fff793b09c4

address of i is 0x7fff793b09c4