Inconsistency for size_t and sizeof

Question

Isn't it bothersome that the result of a native operator cannot be defined without including a header file?

According to this page, size_t is defined in headers cstddef, cstdio, cstring, ctime, and cstdlib. Thus, if neither of those headers is included then size_t should be undefined. However, the following program compiles without any warning (using MSVC 2015RC).

int main() {     auto d_Size = sizeof( int );     return 0; } 

It seems that size_t is somewhat of a bastard between a native type and a typedef. What does the standard say?

Answer 1

5.3.3 Sizeof [expr.sizeof]

1) The sizeof operator yields the number of bytes in the object representation of its operand. The operand is either an expression, which is an unevaluated operand (Clause 5), or a parenthesized type-id. The sizeof operator shall not be applied to an expression that has function or incomplete type, to the parenthesized name of such types, or to a glvalue that designates a bit-field. sizeof(char), sizeof(signed char) and sizeof(unsigned char) are 1. The result of sizeof applied to any other fundamental type (3.9.1) is implementation-defined. [ Note: in particular, sizeof(bool), sizeof(char16_t), sizeof(char32_t), and sizeof(wchar_t) are implementation-defined.75 — end note ] [ Note: See 1.7 for the definition of byte and 3.9 for the definition of object representation. — end note ]

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 ]

However, std::size_t is just a type alias. The sizeof operator can return its result without any need of "accessing" the type alias; the result of sizeof is some fundamental type (implementation defined), which is then aliased as std::size_t in <cstddef>.

Note also that in C++ typedef or using do not define a new type (i.e. a strong type), but only an alias (i.e. their typeid are the same). Hence, in your case, auto just deduces the fundamental type returned by the sizeof operator, which is the same as the type alias std::size_t. No problem for the compiler.

Answer 2

According to the C++ standard, std::size_t is defined in <cstddef>.

5.3.3 Sizeof

...

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 ]