enable_if method specialization

Question

template
struct A
{
    A operator%( const T& x);
};

template
A A::operator%( const T& x ) {          


        

Answer 1

Use overloading instead of explicit specialization when you want to refine the behavior for a more specific parameter type. It's easier to use (less surprises) and more powerful

template
struct A
{
    A operator%( const T& x) { 
      return opModIml(x, std::is_floating_point()); 
    }

    A opModImpl(T const& x, std::false_type) { /* ... */ }
    A opModImpl(T const& x, std::true_type) { /* ... */ }
};

An example that uses SFINAE (enable_if) as you seem to be curious

template
struct A
{
    A operator%( const T& x) { 
      return opModIml(x); 
    }

    template::value>::type>
    A opModImpl(U const& x) { /* ... */ }

    template::value>::type>
    A opModImpl(U const& x) { /* ... */ }
};

Way more ugly of course. There's no reason to use enable_if here, I think. It's overkill.