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Asked: May 11, 20262026-05-11T12:37:09+00:00 2026-05-11T12:37:09+00:00

I’m in the process of designing several classes that need to support operators !=

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I’m in the process of designing several classes that need to support operators !=, >, <=, and >=. These operators will be implemented in terms of operators == and <.

At this stage, I need to make a choice between inheritance¹ and forcing my consumers to use std::rel_ops² ‘manually’.

[1] Inheritance (possible implementation):

template<class T> class RelationalOperatorsImpl {     protected:     RelationalOperatorsImpl() {}     ~RelationalOperatorsImpl() {}      friend bool operator!=(const T& lhs, const T& rhs) {return !(lhs == rhs);}     friend bool operator>(const T& lhs, const T& rhs) {return (rhs < lhs);}     friend bool operator<=(const T& lhs, const T& rhs) {return !(rhs < lhs);}     friend bool operator>=(const T& lhs, const T& rhs) {return !(lhs < rhs);} };  template<typename T> class Foo : RelationalOperatorsImpl< Foo<T> > {     public:     explicit Foo(const T& value) : m_Value(value) {}      friend bool operator==(const Foo& lhs, const Foo& rhs) {return (lhs.m_Value == rhs.m_Value);}     friend bool operator<(const Foo& lhs, const Foo& rhs) {return (lhs.m_Value < rhs.m_Value);}      private:     T m_Value; }; 

[2] std::rel_ops glue:

template<typename T> class Foo {     public:     explicit Foo(const T& value) : m_Value(value) {}      friend bool operator==(const Foo& lhs, const Foo& rhs) {return (lhs.m_Value == rhs.m_Value);}     friend bool operator<(const Foo& lhs, const Foo& rhs) {return (lhs.m_Value < rhs.m_Value);}      private:     T m_Value; };  void Consumer() {     using namespace std::rel_ops;      //Operators !=, >, >=, and <= will be instantiated for Foo<T> (in this case) on demand. } 

I’m basically trying to avoid code repetition. Any thoughts as to which method ‘feels’ better?

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  1. 2026-05-11T12:37:10+00:00Added an answer on May 11, 2026 at 12:37 pm

    Have you considered using boost, and having your class inherit from boost::less_than_comparable<T> and boost::equality_comparable<T>? It is akin to your first suggestion, with some pros and cons. Pros: avoids code duplication; Cons: creates a dependency on boost.

    Since boost is a very common C++ library (if you don’t use it already, you should seriously consider start using it), the con factor is dimmed.

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