Re: [boost] result_of<F()>::type is void ... why?

From: Tobias Schwinger (tschwinger_at_[hidden])
Date: 2007-05-21 09:59:38

Peter Dimov wrote:
> Eric Niebler wrote:
> 
>> I've noticed something odd about the specification of tr1::result_of.
>> In result_of<F()>::type, if F is a class type that doesn't have a
>> nested result_type typedef, the result is "void". This is only true
>> for F(); that is, when there are no arguments.
> 
> This special case is needed in situations such as:
> 
> template<class F> struct X
> {
>     F f_;
> 
>     result_of<F()>::type operator()() { return f_; }
> 
>     template<class A1>
>         result_of<F(A1)>::type operator()(A1& a1) { return f_(a1); }
> 
>     // ...
> };
> 
> where X can be instantiated with non-nullary function objects, or with types 
> that aren't function objects at all, as in:
> 
>     X<int> x;
> 
> The above instantiates the declaration of X::operator()(), which attempts to 
> instantiate result_of<int()>, which would fail without the kludge.
> 
> There's no such problem with the A1 overload since, as a template, it's not 
> instantiated unless used.
> 
> With variadic templates a special case is no longer necessary and I believe 
> that the C++0x result_of no longer has one. 

Thanks.

Will C++0x change the rules for operator overloading to be less 
restrictive (as the above problem would also be solved by a nullary, 
non-member call operator), BTW?

Regards,
Tobias