Merging interfaces, without merging

Question

I was thinking, does C++ or Java have a way to do something like this

Interface IF1{
    ....
};

Interface IF2{
    ....
};


function f(Object o : Implements IF1, IF2){
    ...
}

meaning a typesystem that allows you to require implementation of interfaces.

Answer 1

You can do this in Java:

public <I extends IF1 & IF2> void methodName(I i){

....

}

This way you force I to implement your two interfaces, otherwise it won't even compile.

Answer 2

In C++, we can use std::is_base_of<IF1, Derived>. This has to be used with the actual derived and base type and will be easy to use with the help of tempaltes.

template<typename T>
void f (T obj)
{
  static_assert(is_base_of<IF1,T>::value && is_base_of<IF2,T>::value,
  "Error: Not implementing proper interfaces.");
  ...
}

If T (a derived class) is not implementing IF1 and IF2, then the assertion will fail at compile time.