Print the return type of a function or functor pointer

Question

I am trying to infer the return type of the called type, i.e. pointer to a function or functor.

Earlier, I asked and received an answer showing how to do this for function pointers with hints for how to do this for functors.

Type inference of return type in C ++ 03

Based on this, I have this helper structure, which now works for functors, but no longer works for function pointers.

// functors
template<class T>
struct Return;
{
    typedef typename T::result_type type;
};

// function pointers
template<class R>
struct Return<R (*)()>
{
        typedef R type;
};

This works for interacting functors that provide a result_typetypedef, but not for function pointers. clang gives the following:

error: type 'int (*) (int)' cannot be used before '::' because it has no members

typedef typename T::result_type type;

, SFINAE, Return.

, , .

http://coliru.stacked-crooked.com/a/53f4c8c90787e329

.

"" , function() ++ 98

, std::vector std::vector . , . :

template<typename F, typename T>
typename expr_check<sizeof(declval<F>()(declval<T>())), std::vector<T> >::type
map_vec(F fnc, const std::vector<T>& source)

F , T , F, std::vector<T>. . declval.

, , C. , T F , . :

template<typename F, typename C>
typename expr_check<sizeof(declval<F>()(declval<int>())),
        std::vector<typename Return<F>::type> >::type
map_vec(F fnc, const C& source);

Answer 1

.

pre-++ 11 , .

++ 11, std::result_of.

Answer 2

, Deduplicator.

:

http://coliru.stacked-crooked.com/a/cdb79bbdeedfaa65

, , , Return, .

// functors
template<class T>
struct Return
{
    typedef typename T::result_type type;
};

// function pointers
template<class R, class T>
struct Return<R (*)(T)>
{
        typedef R type;
};

// function pointers
template<class R, class T, class U>
struct Return<R (*)(T, U)>
{
        typedef R type;
};