Adding overload operator == with enable_if

Question

I have a working template class point with an overload for operator==.

Because of floating point comparison, I was trying to add a second overload with enable_if for floating point to use an almost equal function.

This is my attempt:

template<typename T>
class Point2D
{
public:
   Point2D(T x, T y);

   Point2D& operator= (const Point2D& point);
   bool     operator==(const Point2D& point) const;
   bool     operator!=(const Point2D& point) const;
};

template<typename T>
Point2D<T>::Point2D(T x, T y) : x_(x), y_(y)
{
}

template<typename T>
Point2D<T>& Point2D<T>::operator=(const Point2D& point)
{
   if(this != &point)
   {
      x_ = point.x_;
      y_ = point.y_;
   }

   return *this;
}

template<typename T>
bool Point2D<T>::operator==(const Point2D& point) const
{
   return (x_ == point.x_) && (y_ == point.y_);
}

template<typename T>
typename std::enable_if<std::is_floating_point<T>::value, bool>::type
Point2D<T>::operator==(const Point2D& point) const
{
   return Traits::almost_equal(x_, point.x_) &&
          Traits::almost_equal(y_, point.y_);
}

Note:

• This is a semplified example.

• The code actually work without the enable_if overload

• I have to separate declaration and implementation (both in .h), so please refer to the code as is.

The error the compiler gives me is

error: prototype for typename std::enable_if<std::is_floating_point<_Tp>::value, bool>::type Point2D<T>::operator==(const Point2D<T>&) const
  does not match any in class Point2D<T>
  Point2D<T>::operator==(const Point2D& point) const
  ^

error: candidate is: bool Point2D<T>::operator==(const Point2D<T>&)
const bool Point2D<T>::operator==(const Point2D& point) const
      ^

I don't understand what the error is referring to.

Answer 1

Really, just don't do that. Do this instead:

template<typename T>
class Point2D
{
  bool equals(const Point2D& other, std::true_type is_floating_point ) const;
  bool equals(const Point2D& other, std::false_type is_floating_point ) const;
public:
  Point2D(T x, T y);

  Point2D& operator= (const Point2D& point);
  bool     operator==(const Point2D& point) const;
  bool     operator!=(const Point2D& point) const;
};

now:

template<typename T>
bool Point2D<T>::operator==(const Point2D& point) const
{
  return this->equals(point, std::is_floating_point<T>{});
}
template<class T>
bool Point2D<T>::equals(const Point2D& point, std::false_type /*is_floating_point*/ ) const {
  return (x_ == point.x_) && (y_ == point.y_);
}
template<class T>
bool Point2D<T>::equals(const Point2D& point, std::true_type /*is_floating_point*/ ) const {
  return Traits::almost_equal(x_, point.x_) &&
      Traits::almost_equal(y_, point.y_);
}

this is tag dispatching. It is cleaner and easier and takes less time to compile than SFINAE enable-if stuff.

Answer 2

The error is clear - your definition does not match your declaration. You need the exact same enable_if both in your prototype and definition. Also, your enable_if will not work, as it needs to happen during substitution (SFINAE).

To make things more readable, you can define an alias and use trailing return types:

template<typename T>
class Point2D
{
   template <typename U>
   using EnableIfFloat = 
      typename std::enable_if<std::is_floating_point<U>::value, bool>::type;

public:
   Point2D(T x, T y);

   template <typename U = T>
   auto operator==(const Point2D& point) const -> EnableIfFloat<U>&;
   bool operator==(const Point2D& point) const;
   bool operator!=(const Point2D& point) const;
};

Definition:

template<typename T>
template<typename U>
auto Point2D<T>::operator==(const Point2D& point) const -> EnableIfFloat<U>&
{
   return Traits::almost_equal(x_, point.x_) &&
          Traits::almost_equal(y_, point.y_);
}