How to deduce the type of the functor's return value?

Question

I would like to write a template function which accepts 2 values and a functor or a lambda. The function calls the functor with those values and returns the result.

Answer 1

You're nearly there; just use decltype:

template  
auto Apply(T x, T y, Fn fn) -> decltype(fn(x, y))
{
  return fn(x, y);
}

You could use std::result_of (Difference between std::result_of and decltype) but why bother?

template  
typename std::result_of::type Apply(T x, T y, Fn fn)
{
  return fn(x, y);
}

---

Regarding the follow-up question: for a function

auto fn() ->  { return ; }

substituting return-type with decltype() will usually work, but can be error prone. For example, consider:

auto f(char c) -> decltype(std::string() += c) { return std::string() += c; }

Here decltype will yield std::string & and your function will return an lvalue reference to a local! This would have to be changed to:

auto f(char c) -> std::remove_reference::type {
    return std::string() += c;
}

In other cases, could yield a value that is not returnable for reason of being e.g. noncopyable, containing a lambda, etc.