Deducing a function pointer return type

问题

I think code will better illustrate my need:

template <typename F>
struct return_type
{
  typedef ??? type;
};

so that:

return_type<int(*)()>::type -> int
return_type<void(*)(int,int)>::type -> void

I know of decltype and result_of but they need to have arguments passed. I want to deduce the return type of a function pointer from a single template parameter. I cannot add the return type as a parameter, because that's exactly what I want to hide here...

I know there's a solution in boost, but I can't use it, and an attempt to dig it out from boost resulted in a spectacular failure (as it often does).

C++11 solutions welcome (as long as supported in VS2012).

回答1:

If you can use variadic templates (November '12 CTP), this should work:

template <class F>
struct return_type;

template <class R, class... A>
struct return_type<R (*)(A...)>
{
  typedef R type;
};

Live example.

If you can't use variadic templates, you'll have to provide specific specialisations for 0, 1, 2, ... parameters (by hand or preprocessor-generated).

EDIT

As pointed out in the comments, if you want to work with variadic functions as well, you'll have to add one extra partial specialisation (or one for each parameter count in the no-variadic-templates case):

template <class R, class... A>
struct return_type<R (*)(A..., ...)>
{
  typedef R type;
};

来源：https://stackoverflow.com/questions/18695564/deducing-a-function-pointer-return-type