How to use sfinae for selecting constructors?

Question

In template meta programming, one can use SFINAE on the return type to choose a certain template member function, i.e.

template struct          


        

Answer 1

With C++20 you can use the requires keyword

With C++20 you can get rid of SFINAE.

The requires keyword is a simple substitute for enable_if!

Note that the case where otherN == N is a special case, as it falls to the default copy ctor, so if you want to take care of that you have to implement it separately:

template struct A {
   A() {}    

   // handle the case of otherN == N with copy ctor
   explicit A(A const& other) { /* ... */ }

   // handle the case of otherN > N, see the requires below
   template requires (otherN > N)
   explicit A(A const& other) { /* ... */ }

   // handle the case of otherN < N, can add requires or not
   template
   explicit A(A const& other) { /* ... */ }
};

The requires clause gets a constant expression that evaluates to true or false deciding thus whether to consider this method in the overload resolution, if the requires clause is true the method is preferred over another one that has no requires clause, as it is more specialized.

Code: https://godbolt.org/z/RD6pcE