why is sizeof(ptrdiff_t) == sizeof(uintptr_t)

Question

I see several posts (such as size_t vs. uintptr_t) about size_t versus uintptr_t/ptrdiff_t, but none about the relative sizes of these new c99 ptr size types.

example machine: vanilla ubuntu 14lts x64, gcc 4.8:

printf("%zu, %zu, %zu\n", sizeof(uintptr_t), sizeof(intptr_t), sizeof(ptrdiff_t));

prints: "8, 8, 8"

this does not make sense to me, as i would expect the diff type, which must be signed, to require more bits than the unsigned ptr itself.

consider:

NULL - (2^64-1)  /*largest ptr, 64bits of 1's.*/

which being 2's complement negative would not fit in 64bits; hence I would expect ptrdiff_t to be larger than than ptr_t.

[a related question is why is intptr_t the same size as uintptr_t .... although i was comfortable this was possibly just to allow a signed type to contain the representation's bits (eg, using signed arithmetic on a negative ptr would (a) be undefined, and (b) have limited utility as ptrs are by definition "positive")]

thanks!

Answer 1

Firstly, it is clear not what uintptr_t is doing here. The languages (C and C++) do not allow you to subtract just any arbitrary pointer values from each other. Two pointers can only be subtracted if they point into the same object (into the same array object). Otherwise, the behavior is undefined. This means that these two pointers cannot possibly be farther than SIZE_MAX bytes apart. Note: the distance is limited by the range of size_t, not by the range of uintptr_t. In general case uintptr_t can be a larger type than size_t. Nobody in C/C++ ever promised you that you should be able to subtract two pointers located UINTPTR_MAX bytes apart.

(And yes, I know that on flat-memory platforms uintptr_t and size_t are usually the same type, at least by range and representation. But from the language point of view it is incorrect to assume that they always are.)

Your NULL - (2^64-1) (if interpreted as address subtraction) is a clear example of such questionable subtraction. What made you think that you should be able to do that in the first place?

Secondly, after switching from the irrelevant uintptr_t to the much more relevant size_t, one can say that your logic is perfectly valid. sizeof(ptrdiff_t) should be greater than sizeof(size_t) because of an extra bit required to represent the signed result. Nevertheless, however weird it sounds, the language specification does not require ptrdiff_t to be wide enough to accommodate all pointer subtraction results, even if two pointers point to parts of the same object (i.e. they are no farther than SIZE_MAX bytes apart). ptrdiff_t is legally permitted to have the same bit-count as size_t.

This means that a "seemingly valid" pointer subtraction may actually lead to undefined behavior simply because the result is too large. If your implementation allows you to declare a char array of size, say, SIZE_MAX / 3 * 2

char array[SIZE_MAX / 3 * 2]; // This is smaller than `SIZE_MAX`

then subtracting perfectly valid pointers to the end and to the beginning of this array might lead to undefined behavior if ptrdiff_t has the same size as size_t

char *b = array;
char *e = array + sizeof array;

ptrdiff_t distance = e - b; // Undefined behavior!

The authors of these languages decided to opt for this easier solution instead of requiring compilers to implement support for [likely non-native] extra wide signed integer type ptrdiff_t.

Real-life implementations are aware of this potential problem and usually take steps to avoid it. They artificially restrict the size of the largest supported object to make sure that pointer subtraction never overflows. In a typical implementation you will not be able to declare an array larger than PTRDIFF_MAX bytes (which is about SIZE_MAX / 2). E.g. even if SIZE_MAX on your platform is 2⁶⁴-1, the implementation will not let you to declare anything larger than 2⁶³-1 bytes (and real-life restrictions derived from other factors might be even tighter than that). With this restriction in place, any legal pointer subtraction will produce a result that fits into the range of ptrdiff_t.

See also,

• Why is the maximum size of an array “too large”?

Answer 2

The accepted answer is not wrong, but does not offer much insight into why intptr_t, size_t and ptrdiff_t is actually useful, and how to use them. So here it is:

• size_t is basically the type of a size_of expression. It is only required to be able to hold the size of the largest object that you can make, including arrays. So if you can only ever use 64k continues memory, then size_t can be as little as 16 bits, even if you have 64 bit pointers.

• ptrdiff_t is the type of pointer difference, e.g &a - &b. And while it is true that 0 - &a is undefined behavior (as doing almost everything in C/C++), whatever it is, must fit into ptrdiff_t. It is usually the same size as pointers, because that makes the most sense. If ptrdiff_t would be a weird size, pointer arithmetics itself would break.

• intptr_t/uintptr_t has the same size as pointers. They fit into the same int*_t pattern, where * is the size of the int. As with all int*_t/uint*_t types the standard for some reason allows them to be larger then required, but that's very rare.

As a rule of thumb, you can use size_t for sizes and array indices, and use intptr_t/uintptr_t for everything pointer related. Do not use ptrdiff_t.