CODEWITHSUNDEEP
c++c++11language-lawyervoiddecltype
skypjack
skypjack

Reputation: 50568

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?

Upvotes: 31

Views: 5092

Answers (2)

Yakk - Adam Nevraumont
Yakk - Adam Nevraumont

Reputation: 275966

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.

Upvotes: 21

skypjack
skypjack

Reputation: 50568

I'm quoting the comment of @JoachimPileborg that seems to explain it correctly:

I think I figured it out now, decltype needs an expression, and not a type. void() is not actually a type here, but an expression, a C-style cast (just like e.g. int(12.34)) void(void) is not an expression therefore it doesn't work. How the compiler parses different things depends on the context, when it expects a type it parses as a type, when it expects an expression it parses as an expression. sizeof() (with the parentheses) expects first of all a type, otherwise it's parsed as a parenthesized expression.

I'm not looking for credits or reputation.
Anyway, I guess that was an interesting answer in the answer that is worth of a dedicated question for future readers.

Upvotes: 6

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