Question

Why is the Child class’s converter constructor called in the code below?

I mean, it automatically converts Base to Child via the Child converter constructor. The code below compiles, but shouldn’t it not compile since I haven’t provided bool Child::operator!=(Base const&)?

class Base
{
};

class Child : public Base
{
public:
  Child() {}
  Child(Base const& base_) :
    Base(base_)
  {
    std::cout <<"should never called!";
  }
  bool operator!=(Child const&)
  {
    return true;
  }
};

void main()
{
  Base  base;
  Child child;

  if(child != base)
    std::cout << "not equal";
  else
    std::cout << "equal";
}

Answer 1

Because you provided a conversion constructor. If you don’t want the compiler to automatically convert Base objects to Child using the

Child(Base const& base_)

constructor, make it explicit:

explicit Child(Base const& base_)

This way, the constructor will only be called when you explicitly specify it, in contexts like:

Child ch(base);
foo(Child(base));
new Child(base);