template<typename T>
struct A
{
A<T> operator%( const T& x);
};
template<typename T>
A<T> A<T>::operator%( const T& x ) { ... }
How can I use enable_if to make the following specialization happen for any floating point type (is_floating_point)?
template<>
A<float> A<float>::operator%( const float& x ) { ... }
EDIT: Here's an answer I came up which is different from the ones posted below...
template<typename T>
struct A
{
T x;
A( const T& _x ) : x(_x) {}
template<typename Q>
typename std::enable_if<std::is_same<Q, T>::value && std::is_floating_point<Q>::value, A<T> >::type operator% ( const Q& right ) const
{
return A<T>(fmod(x, right));
}
template<typename Q>
typename std::enable_if<std::is_convertible<Q, T>::value && !std::is_floating_point<Q>::value, A<T> >::type operator% ( const Q& right ) const
{
return A<T>(x%right);
}
};
Like the below posters say, using enable_if may not be ideal for this problem (it's very difficult to read)
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<typename T>
struct A
{
A<T> operator%( const T& x) {
return opModIml(x, std::is_floating_point<T>());
}
A<T> opModImpl(T const& x, std::false_type) { /* ... */ }
A<T> opModImpl(T const& x, std::true_type) { /* ... */ }
};
An example that uses SFINAE (enable_if
) as you seem to be curious
template<typename T>
struct A
{
A<T> operator%( const T& x) {
return opModIml(x);
}
template<typename U,
typename = typename
std::enable_if<!std::is_floating_point<U>::value>::type>
A<T> opModImpl(U const& x) { /* ... */ }
template<typename U,
typename = typename
std::enable_if<std::is_floating_point<U>::value>::type>
A<T> opModImpl(U const& x) { /* ... */ }
};
Way more ugly of course. There's no reason to use enable_if
here, I think. It's overkill.