What Is The Difference Between %p And 0x%08x - Stack Overflow

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Learn more about Labs What is the difference between %p and 0x%08x Ask Question Asked 9 years, 1 month ago Modified 9 years, 1 month ago Viewed 14k times 3

the question says it all. In c, when trying to get the address of a pointer or something, I know the commands:

%p

and

0x%08x

I know that they both return addresses, but why use one over another?

Check this out:

int main () { int i[1] = {1}; int *int_pointer; int_pointer = i; printf("0x%08x\n",&i); printf("0x%08x\n", int_pointer); printf("%p\n",&i); printf("%p\n", int_pointer); }

This code yields this response: 0x60506b00 0x60506b00 0x7ffe60506b00 0x7ffe60506b00

if both refer to the addresses, then how come they are different depending on if you use %p or 0x%08x?

Which is the actual address?

• c

Share Improve this question Follow edited Oct 25, 2015 at 0:39 yasgur99 asked Oct 25, 2015 at 0:02 yasgur99yasgur99 83822 gold badges1212 silver badges3737 bronze badges 1

• 2 Your pointers are 64 bit, and %08x only prints 32 bits of hex. – JS1 Commented Oct 25, 2015 at 2:43

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2 Answers 2

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printf("%p", p) expects a void* pointer as argument. printf("0x%08x", n) expects an unsigned int as argument.

Passing a pointer to the latter might work anyway provided that an unsigned int and a pointer are "similar enough" in their representations on the stack. But to be safe and correct no matter what your machine is, always use "%p" with pointers.

Share Improve this answer Follow answered Oct 25, 2015 at 0:08 aschepleraschepler 72.1k1010 gold badges112112 silver badges170170 bronze badges 5

• so if i was using a char, I would just convert it into an int. Correct me if im wrong – yasgur99 Commented Oct 25, 2015 at 0:15
• i edited my original post to reflect an example of the usage and i'm still confused. do you mind clarifying it? – yasgur99 Commented Oct 25, 2015 at 1:29
• @yasgur99: it appears that pointers are 64-bits wide on your machine, but the %08x conversion specifier only prints the low-order 32 bits. Pointers are not unsigned integers, and they don't have to have the same size or representation as unsigned integers. Always use %p to print pointer values, at least if you want meaningful output. – John Bode Commented Oct 25, 2015 at 4:26
• so what should I use if I wanted to print the address of integers or other variables – yasgur99 Commented Oct 25, 2015 at 4:28
• @yasgur99: assuming you want to print the address of a variable named a (regardless of type), you would write printf( "%p\n", (void *) &a ); – John Bode Commented Oct 26, 2015 at 16:59

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Because %p is the portable way of doing it. Even though a pointer stores an address, it is not an integer (and the standard does not require a compiler to use integers for pointer representations), hence it is not appropriate to use an integer specifier.

Share Improve this answer Follow answered Oct 25, 2015 at 0:08 vsoftcovsoftco 56.4k1212 gold badges147147 silver badges263263 bronze badges 2

• what about when using a char – yasgur99 Commented Oct 25, 2015 at 0:10
• a char is implicitly convertible to an integer. – vsoftco Commented Oct 25, 2015 at 0:10

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