enable_if method specialization

Question

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

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


        

Answer 1

You can also use a default boolean template parameter like this:

template
struct A
{
    T x;

    A( const T& _x ) : x(_x) {}

    template
    typename std::enable_if::value && EnableBool, A >::type 
    operator% ( const T& right ) const
    {
        return A(fmod(x, right));
    }

    template
    typename std::enable_if::value && EnableBool, A >::type 
    operator% ( const T& right ) const
    {
        return A(x%right);
    }
};