change function defination based on type trait?

Question

I would like to change my function definition based on type by using enable_if_t. Some thing similar to this:

#include<type_traits>
template<typename T> struct A;

template<typename T>
struct A{
    std::enable_if_t<std::is_arithmetic<T>::value, bool>
    test()
    {
        return true;
    }
    std::enable_if_t<!std::is_arithmetic<T>::value,  bool>
    test()
    {
        return false;    
    }

};

Currently it cannot compile at all.

Class methods with the same name and signature are not allowed. A function's return type is not a part of its signature. The easiest solution here is to declare the return type as `auto`, and use `if constexpr` to return the appropriate type. — Sam Varshavchik, Jun 09 '20 at 17:32

score 0 · Accepted Answer · answered Jun 09 '20 at 17:33

You can try with

template <typename U = T>
std::enable_if_t<std::is_arithmetic<U>::value, bool>
test()
 { return true; }

template <typename U = T>
std::enable_if_t<!std::is_arithmetic<U>::value,  bool>
test()
 { return false; }

I mean... SFINAE works over template parameter of the functions/methods, not over template parameters of the class containing the methods.

With typename U = T you transform T, the class template parameter, in U, a method template parameter.

change function defination based on type trait?

1 Answers1