c++ - Does `sizeof` *really* evaluate to a `std::size_t`? Can it?

Question

Take the following standard passage:

[C++11: 5.3.3/6]: The result of sizeof and sizeof... is a constant of type std::size_t. [ Note: std::size_t is defined in the standard header <cstddef> (18.2). —end note ]

Now:

[C++11: 18.2/6]: The type size_t is an implementation-defined unsigned integer type that is large enough to contain the size in bytes of any object.

Granted, the passage doesn't require that size_t is a type alias defined with typedef, but since it's explicitly stated to be made available by the standard header <cstddef>, I think we can take as read that failing to include <cstddef> should remove any guarantee that size_t shall be available to a program.

However, according to that first quote, we can regardless obtain an expression of type std::size_t.

We can actually demonstrate both of these facts:

int main()
{
    typedef decltype(sizeof(0)) my_size_t;

    my_size_t x   = 0;  // OK
    std::size_t y = 1;  // error: 'size_t' is not a member of 'std'
}

std::size_t is not visible to the program, but sizeof(0) still gives us one? Really?

Is it therefore not correct to say that 5.3.3/6 is flawed, and that it actually has "the same type as whatever std::size_t resolves to", but not std::size_t itself?

Sure, the two are one and the same if std::size_t is a type alias but, again, nowhere is this actually required.

Answer 1

The standard just mandates that the type of sizeof(expr) is the same type as std::size_t. There is no mandate that using sizeof(expr) makes the name std::size_t available and since std::size_t just names one of the the built-in integral types there isn't really a problem.