Can sizeof(size_t) be less than sizeof(int)?

Question

Can sizeof(size_t) be less than sizeof(int)?

Do the C and/or C++ standards guarantee that using unsigned int for array indexing is always safe?

Answer 1

Yes, sizeof(size_t) can, in principle, be less than sizeof(int). I don't know of any implementations where this is true, and it's likely that there are none. I can imagine an implementation with 64-bit int and 32-bit size_t.

But indexing an array with unsigned int is safe -- as long as the value of the index is within the bounds imposed by the length of the array. The argument to the [] operator is merely required to be an integer. It's not converted to size_t. It's defined in terms of pointer arithmetic, in which the + operator has one argument that's a pointer and another argument that is of any integer type.

If unsigned int is wider than size_t, then an unsigned int index value that exceeds SIZE_MAX will almost certainly cause problems because the array isn't that big. In C++14 and later, defining a type bigger than SIZE_MAX bytes is explicitly prohibited (3.9.2 [compound.types] paragraph 2; section 6.9.2 in C++17). In earlier versions of C++, and in all versions of C, it isn't explicitly prohibited, but it's unlikely that any sane implementation would allow it.