CODEWITHSUNDEEP
c++templatesc++11
blue
blue

Reputation: 2793

using std::result_of to determine the return type of a template argument

I think the snippet of code is self explanatory, but basically the template function ExecFunc should be able to execute another function and return its result. I know I can achieve similar results using decltype instead of result_of, but this question is to understand why what I've written does not work: the snippet does not compile on gcc v4.9.2.

This is what I have:

#include <type_traits>

int f(int i)
{
   return i;
}

template<class F, class T>
auto ExecFunc(F f, T arg) -> typename std::result_of<F()>::type
{
  return f(arg);
}

int main() {
   auto a = ExecFunc(f, 3);
   return 0;
}

and this is the compiler output:

prova.cpp: In function ‘int main()’:
prova.cpp:15:26: error: no matching function for call to ‘ExecFunc(int (&)(int), int)’
auto a = ExecFunc(f, 3);
                      ^
prova.cpp:15:26: note: candidate is:
prova.cpp:9:6: note: template<class F, class T> typename std::result_of<F()>::type ExecFunc(F, T)
 auto ExecFunc(F f, T arg) -> typename std::result_of<F()>::type
      ^
prova.cpp:9:6: note:   template argument deduction/substitution failed:
prova.cpp: In substitution of ‘template<class F, class T> typename std::result_of<F()>::type ExecFunc(F, T) [with F = int (*)(int); T = int]’:
prova.cpp:15:26:   required from here
prova.cpp:9:6: error: no type named ‘type’ in ‘class std::result_of<int (*())(int)>’

N.B. this question might look like a duplicate of this one but the accepted solution doesn't work for me (at least, as far as I can tell I have incorporated the solution in my code).

Upvotes: 4

Views: 2332

Answers (3)

Aaron McDaid
Aaron McDaid

Reputation: 27133

This is the perfect time to use decltype

template<class F, class T>
auto ExecFunc(F f, T arg) -> decltype(f(arg))

Upvotes: 0

NathanOliver
NathanOliver

Reputation: 180594

The function you have is int f(int i) but you are calling F() which is unknown. std::result_of<F()>::type should be std::result_of<F(T)>::type.

Live Example

Upvotes: 4

vukung
vukung

Reputation: 1884

The problem is with the parameter of result_of, it should be:

-> typename std::result_of<F(T)>::type

Upvotes: 2

Related Questions