size_t vs. uintptr_t

Question

The C standard guarantees that size_t is a type that can hold any array index. This means that, logically, size_t should be able to hold any pointer type. I've read on some sites that I found on the Googles that this is legal and/or should always work:

void *v = malloc(10);
size_t s = (size_t) v;

So then in C99, the standard introduced the intptr_t and uintptr_t types, which are signed and unsigned types guaranteed to be able to hold pointers:

uintptr_t p = (size_t) v;

So what is the difference between using size_t and uintptr_t? Both are unsigned, and both should be able to hold any pointer type, so they seem functionally identical. Is there any real compelling reason to use uintptr_t (or better yet, a void *) rather than a size_t, other than clarity? In an opaque structure, where the field will be handled only by internal functions, is there any reason not to do this?

By the same token, ptrdiff_t has been a signed type capable of holding pointer differences, and therefore capable of holding most any pointer, so how is it distinct from intptr_t?

Aren't all of these types basically serving trivially different versions of the same function? If not, why? What can't I do with one of them that I can't do with another? If so, why did C99 add two essentially superfluous types to the language?

I'm willing to disregard function pointers, as they don't apply to the current problem, but feel free to mention them, as I have a sneaking suspicion they will be central to the "correct" answer.

user_2410359 · Answer

size_t vs. uintptr_t

In addition to other good answers:

size_t is defined in , , , , , , . It is at least 16-bit.

uintptr_t is defined in . It is optional. A compliant library might not define it, likely because there is not a wide-enough integer type to round trip a void*-uintptr_t-void *.

Both are unsigned integer types.

Note: the optional companion intptr_t is a signed integer type.

user_27491 · Answer

int main(){
  int a[4]={0,1,5,3};
  int a0 = a[0];
  int a1 = *(a+1);
  int a2 = *(2+a);
  int a3 = 3[a];
  return a2;
}

Implying that intptr_t must always substitute for size_t and visa versa.

size_t vs. uintptr_t

8 Answers