Vincent
Vincent

Reputation: 60381

Get the result type of a function in c++11

Consider the following function in C++11:

template<class Function, class... Args, typename ReturnType = /*SOMETHING*/> 
inline ReturnType apply(Function&& f, const Args&... args);

I want ReturnType to be equal to the result type of f(args...) What do I have to write instead of /*SOMETHING*/ ?

Upvotes: 4

Views: 2864

Answers (3)

Sarfaraz Nawaz
Sarfaraz Nawaz

Reputation: 361422

I think you should rewrite your function template using trailing-return-type as:

template<class Function, class... Args> 
inline auto apply(Function&& f, const Args&... args) -> decltype(f(args...))
{
    typedef decltype(f(args...)) ReturnType;

    //your code; you can use the above typedef.
}

Note that if you pass args as Args&&... instead of const Args&..., then it is better to use std::forward in f as:

decltype(f(std::forward<Args>(args)...))

When you use const Args&..., then std::forward doesn't make much sense (at least to me).

It is better to pass args as Args&&... called universal-reference and use std::forward with it.

Upvotes: 15

lawilog
lawilog

Reputation: 171

There are situations where std::result_of is more useful. For example, say you want to pass this function:

int ff(int& out, int in);

and inside apply() call it like so

int res;
f(res, args...);

then I wouldn't know how to use decltype, because I have no int lvalue reference at hand. With result_of, you don't need variables:

template<class Function, class... Args> 
typename std::result_of<Function(const Args&...)>::type apply(Function&& f, const Args&... args)
{
  typedef typename std::result_of<F(const Args&...)>::type ReturnType;

  // your code
}

Upvotes: 0

Ben Voigt
Ben Voigt

Reputation: 283634

It doesn't need to be a template parameter, since it isn't used for overload resolution. Try

template<class Function, class... Args> 
inline auto apply(Function&& f, const Args&... args) -> decltype(f(std::forward<const Args &>(args)...));

Upvotes: 4

Related Questions