Where Is '%P' Useful with Printf

Where is `%p` useful with printf?

They do not do the same thing. The latter printf statement interprets b as an unsigned int, which is wrong, as b is a pointer.

Pointers and unsigned ints are not always the same size, so these are not interchangeable. When they aren't the same size (an increasingly common case, as 64-bit CPUs and operating systems become more common), %x will only print half of the address. On a Mac (and probably some other systems), that will ruin the address; the output will be wrong.

Always use %p for pointers.

Usage of %p in printf ?

This is leveraging undefined behavior.

By deliberately not providing values to printf, va_arg is going to be pulling arbitrary values off of the stack to be printed. It is not correct code. And in fact it looks like this code snippet is trying to explain hacking techniques, which often take advantage of glitches that occur when undefined behavior happens.

The %p printf parameter

Trash. printf uses a variable-length argument list. It uses the format string to determine how many arguments you actually passed. If you did not actually pass anything in, it will still read from basically arbitrary portions of memory as though you did. The result is undefined/trash.

Some compilers will be able to catch this situation with a warning because the printf family of functions is so popular. Some cases may crash your system if the function tries to read from memory you do not have access to. There is no way to tell how it will behave next time even if you have obtained a certain result.

How to printf a memory address in C

Use the format specifier %p:

printf("variable A is at address: %p\n", (void*)&A);

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The standard requires that the argument is of type void* for %p specifier. Since, printf is a variadic function, there's no implicit conversion to void * from T * which would happen implicitly for any non-variadic functions in C. Hence, the cast is required. To quote the standard:

7.21.6 Formatted input/output functions (C11 draft)

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.

Whereas you are using %x, which expects unsigned int whereas &A is of type int *. You can read about format specifiers for printf from the manual. Format specifier mismatch in printf leads to undefined behaviour.

Correct format specifier to print pointer or address?

The simplest answer, assuming you don't mind the vagaries and variations in format between different platforms, is the standard %p notation.

The C99 standard (ISO/IEC 9899:1999) says in §7.19.6.1 ¶8:

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.

(In C11 — ISO/IEC 9899:2011 — the information is in §7.21.6.1 ¶8.)

On some platforms, that will include a leading 0x and on others it won't, and the letters could be in lower-case or upper-case, and the C standard doesn't even define that it shall be hexadecimal output though I know of no implementation where it is not.

It is somewhat open to debate whether you should explicitly convert the pointers with a (void *) cast. It is being explicit, which is usually good (so it is what I do), and the standard says 'the argument shall be a pointer to void'. On most machines, you would get away with omitting an explicit cast. However, it would matter on a machine where the bit representation of a char * address for a given memory location is different from the 'anything else pointer' address for the same memory location. This would be a word-addressed, instead of byte-addressed, machine. Such machines are not common (probably not available) these days, but the first machine I worked on after university was one such (ICL Perq).

If you aren't happy with the implementation-defined behaviour of %p, then use C99 <inttypes.h> and uintptr_t instead:

printf("0x%" PRIXPTR "\n", (uintptr_t)your_pointer);

This allows you to fine-tune the representation to suit yourself. I chose to have the hex digits in upper-case so that the number is uniformly the same height and the characteristic dip at the start of 0xA1B2CDEF appears thus, not like 0xa1b2cdef which dips up and down along the number too. Your choice though, within very broad limits. The (uintptr_t) cast is unambiguously recommended by GCC when it can read the format string at compile time. I think it is correct to request the cast, though I'm sure there are some who would ignore the warning and get away with it most of the time.

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Kerrek asks in the comments:

I'm a bit confused about standard promotions and variadic arguments. Do all pointers get standard-promoted to void*? Otherwise, if int* were, say, two bytes, and void* were 4 bytes, then it'd clearly be an error to read four bytes from the argument, non?

I was under the illusion that the C standard says that all object pointers must be the same size, so void * and int * cannot be different sizes. However, what I think is the relevant section of the C99 standard is not so emphatic (though I don't know of an implementation where what I suggested is true is actually false):

§6.2.5 Types

¶26 A pointer to void shall have the same representation and alignment requirements as a pointer to a character type.³⁹⁾ Similarly, pointers to qualified or unqualified versions of compatible types shall have the same representation and alignment requirements. All pointers to structure types shall have the same representation and alignment requirements as each other. All pointers to union types shall have the same representation and alignment requirements as each other. Pointers to other types need not have the same representation or alignment requirements.

³⁹⁾ The same representation and alignment requirements are meant to imply interchangeability as arguments to functions, return values from functions, and members of unions.

(C11 says exactly the same in the section §6.2.5, ¶28, and footnote 48.)

So, all pointers to structures must be the same size as each other, and must share the same alignment requirements, even though the structures the pointers point at may have different alignment requirements. Similarly for unions. Character pointers and void pointers must have the same size and alignment requirements. Pointers to variations on int (meaning unsigned int and signed int) must have the same size and alignment requirements as each other; similarly for other types. But the C standard doesn't formally say that sizeof(int *) == sizeof(void *). Oh well, SO is good for making you inspect your assumptions.

The C standard definitively does not require function pointers to be the same size as object pointers. That was necessary not to break the different memory models on DOS-like systems. There you could have 16-bit data pointers but 32-bit function pointers, or vice versa. This is why the C standard does not mandate that function pointers can be converted to object pointers and vice versa.

Fortunately (for programmers targetting POSIX), POSIX steps into the breach and does mandate that function pointers and data pointers are the same size:

§2.12.3 Pointer Types

All function pointer types shall have the same representation as the type pointer to void. Conversion of a function pointer to void * shall not alter the representation. A void * value resulting from such a conversion can be converted back to the original function pointer type, using an explicit cast, without loss of information.

Note:
The ISO C standard does not require this, but it is required for POSIX conformance.

So, it does seem that explicit casts to void * are strongly advisable for maximum reliability in the code when passing a pointer to a variadic function such as printf(). On POSIX systems, it is safe to cast a function pointer to a void pointer for printing. On other systems, it is not necessarily safe to do that, nor is it necessarily safe to pass pointers other than void * without a cast.

%p Format specifier in c

If this is what you are asking, %p and %Fp print out a pointer, specifically the address to which the pointer refers, and since it is printing out a part of your computer's architecture, it does so in Hexadecimal.

In C, you can cast between a pointer and an int, since a pointer is just a 32-bit or 64-bit number (depending on machine architecture) referring to the aforementioned chunk of memory.

And of course, 55 in hex is 85 in decimal.

printf format specifier flags '0' for %p, correct or not?

%p is for pointer address, which is an integer number after all

Although pointers do have numeric representation, the standard does not consider pointers an integral data type. Therefore, using %08p format causes undefined behavior.

You can work around this problem by using uintptr_t data type, converting your pointer to it, and then printing it as an unsigned integer:

#include <cinttypes>
#include <cstdint>

int main() {
    int i = 123;
    void* p = &i;
    printf("%08" PRIxPTR "\n", (uintptr_t)p);
    return 0;
}

Demo.

What's the proper use of printf to display pointers padded with 0s

#include <inttypes.h>

#include <stdint.h>

printf("%016" PRIxPTR "\n", (uintptr_t)ptr);

but it won't print the pointer in the implementation defined way (says DEAD:BEEF for 8086 segmented mode).

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