Return type of function call on template argument class

Question

I am frequently finding myself dealing with templates and return values of functions in said templates. And I invariably end up with something like this:

template<typename Data>
auto foo(const Data& d) ->
    typename std::decay<decltype(reinterpret_cast<const Data*>(0)->operator()(0, 0))>::type
{
    typedef typename std::decay<decltype(reinterpret_cast<const Data*>(0)->operator()(0, 0))>::type return_t;
    ...
}

While this works, it is ugly and not really obvious what I want from a quick look. Is there a more readable, less 'hackish' way to get "the return type of calling this method on the template argument" ?

Answer 1

Without alias templates c++11:

template <typename Data>
auto foo(const Data& d) -> typename std::decay<decltype(d(0, 0))>::type
{
    return {};
}  

or:

template <typename Data
        , typename return_r = typename std::decay<
              typename std::result_of<const Data(int, int)>::type
          >::type>
return_r foo(const Data& d)
{
    return {};
}

---

With alias templates (c++14 or written by hand):

template <typename Data>
auto foo(const Data& d) -> std::decay_t<decltype(d(0, 0))>
{
    return {};
}

or:

template <typename Data
        , typename return_r = std::decay_t<std::result_of_t<const Data(int, int)>>>
return_r foo(const Data& d)
{
    return {};
}

Use std::invoke_result_t in c++17.