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c++c++11language-lawyervoiddecltype

Is the void() in decltype(void()) an expression or is it a function type?

This is a follow-up of this question, more precisely of the comments of this answer.

What does the void() in decltype(void()) represent exactly?
Does it represent a function type, an expression or whatever?

Solution

  • Using a hyperlinked C++ grammar, the parsing of decltype(void()) is:

    decltype( expression )
decltype( assignment-expression )
decltype( conditional-expression )

    ... lots of steps involving order of operations go here ...

    decltype( postfix-expression )
decltype( simple-type-specifier ( expression-listopt ) )
decltype( void() )

    So void() is a kind of expression here, in particular a postfix-expression.

    Specifically, quoting section 5.2.3 [expr.type.conf] paragraph 2 of the 2011 ISO C++ standard:

    The expression T(), where T is a simple-type-specifier or typename-specifier for a non-array complete object type or the (possibly cv-qualified) void type, creates a prvalue of the specified type, which is value-initialized (8.5; no initialization is done for the void() case).

    So void() is an expression of type void, just as int() is an expression of type int (with value 0). Clearly a void expression has no value, but here it's the operand of decltype, so it's not evaluated. decltype refers only to its operand's type, not its value.

    decltype(void()) is simply a verbose way of referring to the type void.