Is it okay to remove const qualifier when the call is from the same non-const version overloaded member function?

Question

For example:

struct B{};

struct A {

 const B& findB() const { /* some non trivial code */ }

 // B& findB() { /* the same non trivial code */ }

 B& findB() { 
       const A& a = *this;
       const B& b = a.findB();
       return const_cast<B&>(b);
  }
};

The thing is I want to avoid repeating the same logic inside the constant findB and non-constant findB member function.

Answer 1

Yes, you can cast the object to const, call the const version, then cast the result to non-const:

return const_cast<B&>(static_cast<const A*>(this)->findB());

Casting away const is safe only when the object in question was not originally declared const. Since you are in a non-const member function, you can know this to be the case, but it depends on the implementation. Consider:

class A {
public:

    A(int value) : value(value) {}

    // Safe: const int -> const int&
    const int& get() const {
        return value;
    }

    // Clearly unsafe: const int -> int&
    int& get() {
        return const_cast<int&>(static_cast<const A*>(this)->get());
    }

private:
    const int value;
};

Generally speaking, my member functions are short, so the repetition is tolerable. You can sometimes factor the implementation into a private template member function and call that from both versions.