Question

using an interface (abstract class) on c++ i have the need to force any class than inherits the interface to implement the operator ==.
consider this situation:

class IBase
{
   virtual void someFunc() const = 0; 
}

class CInheritClass : public IBase
{
   virtual void someFunc() const; 
   virtual bool operator== ( const CInheritClass& obj ) const;
}

void main()
{
  CInheritClass instance;
}

class CInheritClass must implement someFunc since it inherits Ibase, implementing virtual bool operator== ( const CInheritClass& obj ) const; is not mandatory.
i would like to modify IBase class in a way that any inheritor X will have to implement

virtual bool operator== ( const X& obj ) const

the following code will works:

template<class X>
class IBase
{
   virtual void someFunc() const = 0; 
   virtual bool operator== ( const X& obj ) const = 0;

}

class CInheritClass : public IBase<CInheritClass>
{
   virtual void someFunc() const; 
   virtual bool operator== ( const CInheritClass& obj ) const;
}

but i am after a solution that does not use templates cause every class that wishes to implement IBase must inherit IBase with itself as the template class class X : public IBase<X> and that is confusing and unclear to any future developer that might have a look on my code.
any idea ?

Answer 1

Use a pure virtual function declaration

virtual bool operator== ( const IBase& obj ) const = 0;

A example of implementation :

bool CInheritClass ::operator==( const IBase& obj ) const
{
     const CInheritClass *o = dynamic_cast<const CInheritClass*>(&obj);
     if (o == NULL) return false;
     // TODO
}

The input type should be the base class, if not you will not be able to do polymorphically call comparison on base objects

For the idea of inherits from a template: warning, the base class is no longer the same between 2 derived class.
=> The purpose of the base class changed : With template you just share code, force user to implement function and prevent duplicate code between the 2 derived class

It really depend what you are trying to do. It can be a good design, or not…