Template class implementing comparison operators - c++

Question

It is a frequent task of mine to write all the overloaded comparison operators to a class, so I've written a template class which implements <,<=,>=,!= if the derived class implements == and <. It is working but features a lot of cast and the not that obvious "Curiously recurring template pattern", so I wonder if are there simpler solutions?
template <class Derived>
class Comparable
{
public:
bool operator!=(const Comparable<Derived>& other) {
return !(static_cast<Derived*>(this)->operator==
(*static_cast<const Derived*>(&other)));
}
bool operator<=(const Comparable<Derived>& other) {
return (static_cast<Derived*>(this)->operator==
(*static_cast<const Derived*>(&other)))
|| (static_cast<Derived*>(this)->operator<
(*static_cast<const Derived*>(&other)));
}
bool operator>(const Comparable<Derived>& other) {
return !(static_cast<Derived*>(this)->operator==
(*static_cast<const Derived*>(&other)))
&& !(static_cast<Derived*>(this)->operator<
(*static_cast<const Derived*>(&other)));
}
bool operator>=(const Comparable<Derived>& other) {
return !(static_cast<Derived*>(this)->operator<
(*static_cast<const Derived*>(&other)));
}
};

Answer 1

In case it is not obvious from the description in the comment:
template <typename T>
struct Comparable {
friend bool operator!=(T const & lhs, T const & rhs) { return !(lhs == rhs); }
friend bool operator> (T const & lhs, T const & rhs) { return rhs < lhs; }
// ...
};
class MyType : Comparable<MyType> {
int data;
friend bool operator==(MyType const & lhs, MyType const & rhs) {
return lhs.data == rhs.data;
}
friend bool operator< (MyType const & lhs, MyType const & rhs) {
return lhs.data < rhs.data;
}
public:
// ...
};
When the compiler encounters MyType a, b; a > b; lookup for the operator will end up doing ADL which will look inside MyType and Comparable<MyType> (as this is a base), where it will find the implementation you need: bool operator>(MyType const&, MyType const&).
The operators being free functions allows for a definition that is outside of the type that is being compared (in this case the base), while making those operators only available through ADL (one of the two arguments must be Comparable<MyType>). The use of a free function also provides type-symmetry, the compiler will allow implicit conversions on both sides, where in the case of a member function it would only allow conversions on the right hand side of the operator.
For completeness, a different trick that can be done is to provide the operators as templates in a namespace together with a tag that can be used to bring that namespace in for ADL purposes:
namespace operators {
template <typename T>
bool operator>(T const & lhs, T const & rhs) {
return rhs < lhs;
}
// rest of the operators come here
struct tag {};
}
class MyType : operators::tag {
int data;
friend bool operator<(T const & lhs, T const & rhs) {
return lhs.data < rhs.data;
}
//...
};
The trick is basically the same, except that in this case the operators are not found inside the base, but in a namespace that is associated with it. This solution is a bit less nice than the previous one, as it is open to different forms of misuse, including using namespace operators; that would make the templated operators available for all types.