Re: [boost] [result_of] now uses decltype on release branch

Subject: Re: [boost] [result_of] now uses decltype on release branch
From: Michel Morin (mimomorin_at_[hidden])
Date: 2012-09-10 05:05:23

Eric Niebler wrote:
>>   template < class , class = void >
>>   struct result_of_aux
>>   {};
>>
>>   template < class F, class ...Args>
>>   struct result_of_aux<F(Args...),
>> decltype(std::declval<F>()(std::declval<Args>()...), void ())>
>>   {
>>       typedef decltype(std::declval<F>()(std::declval<Args>()...)) type;
>>   };
>
> What is void()? That's not any expression I recognize. It's a function
> type.

5.2.3 [expr.type.conv] p2
  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).

> Anyway, keep in mind that std::declval<F>()(...) could have an
> incomplete type, so you can't really use it in an expression.
> You should use it in decltype,

All the use of `std::declval<F>()(...)` is in decltype
and so I don't see any problem in the code.
Do you mean
    struct result_of_aux<
        F(Args...)
      , decltype(std::declval<F>()(std::declval<Args>()...), void ())
    >
can be a hard error?

Regards,
Michel