c - printf("%p") and casting to (void *)

Question

In a recent question, someone mentioned that when printing a pointer value with printf, the caller must cast the pointer to void *, like so:

int *my_ptr = ....

printf("My pointer is: %p", (void *)my_ptr);

For the life of me I can't figure out why. I found this question, which is almost the same. The answer to question is correct - it explains that ints and pointers are not necessarily the same length.

This is, of course, true, but when I already have a pointer, like in the case above, why should I cast from int * to void *? When is an int * different from a void *? In fact, when does (void *)my_ptr generate any machine code that's different from simply my_ptr?

UPDATE: Multiple knowledgeable responders quoted the standard, saying passing the wrong type may result in undefined behavior. How? I expect printf("%p", (int *)ptr) and printf("%p", (void *)ptr) to generate the exact same stack-frame. When will the two calls generate different stack frames?

Answer 1

The %p conversion specifier requires an argument of type void *. If you don't pass an argument of type void *, the function call invokes undefined behavior.

From the C Standard (C11, 7.21.6.1p8 Formatted input/output functions):

"p - The argument shall be a pointer to void."

Pointer types in C are not required to have the same size or the same representation.

An example of an implementation with different pointer types representation is Cray PVP where the representation of pointer types is 64-bit for void * and char * but 32-bit for the other pointer types.

See "Cray C/C++ Reference Manual", Table 3. in "9.1.2.2" http://docs.cray.com/books/004-2179-003/004-2179-003-manual.pdf