C++ get return type for template argument

Question

I have

template<typename DistanceFunc>
class C
{
    using DistanceType = // the return type from DistanceFunc

public:
    C(DistanceType distance, DistanceFunc f)
    : m_distance(distance),
      m_f(f)
    {}

private:
    DistanceType m_distance;
    DistanceFunc m_f;
};

How do I derive the return type from DistanceFunc for DistanceType?

Answer 1

How do I derive the return type from DistanceFunc for DistanceType?

Another solution could be use template specialization.

You can declare C as receiving a typename, without defining it

template <typename>
class C;

Then you can define a specialization of C receiving a (pointer to) function type as follows

template <typename DistanceType, typename ... Args>
class C<DistanceType(*)(Args...)>
 {
   using DistanceFunc = DistanceType(*)(Args...);

public:
    C(DistanceType distance, DistanceFunc f)
    : m_distance(distance), m_f(f)
    {}

private:
    DistanceType m_distance;
    DistanceFunc m_f;
};

As you can see the function type is resolved as a return type and the types of the arguments. Now DistanceType is simply deduced (as return type) and you have to recreate DistanceFunc.

You can use it as follows

int foo (int, long)
 { return 0; }

// ...

C<decltype(&foo)> c0{0, &foo};

Starting from C++17 you can also add a deduction guide

template <typename DistanceType, typename ... Args>
C(DistanceType, DistanceType(*)(Args...)) -> C<DistanceType(*)(Args...)>;

so you can declare a C object simply as

C c0{0, &foo}; // C<decltype(&foo)> is deduced

Answer 2

How do I derive the return type from DistanceFunc for DistanceType?

Depends. How is called a function of type DistanceFunc?

Suppose is called with an int, you can try something as

using DistanceType = decltype(std::declval<DistanceFunc>()(std::declval<int>()));

If you have the types received from DistanceFunc, using decltype() and std::declval() you can simulate a call to the function and obtain the resulting type.