Question

I always think I know C++ pretty well, but sometimes I’m surprised by even the most fundamental things.

In the following scenario, I’m confused as to why the constructor Derived::Derived(const Base&) is invoked:

class Base
{ };

class Derived : public Base
{
    public:

    Derived() { }

    Derived(const Base& b) 
    {
        std::cout << "Called Derived::Derived(const Base& b)" << std::endl;
    }
};

int main()
{
    Derived d;
    Base b;
    d = b;
}

This outputs: Called Derived::Derived(const Base& b), indicating that the second constructor in Derived was invoked. Now, I thought I knew C++ pretty well, but I can’t figure out why that constructor would be invoked. I understand the whole “rule of four” concept, and I would think that the expression d = b would do one of two things: Either it would 1) invoke the implicit (compiler-generated) assignment operator of Base, or 2) Trigger a compiler error complaining that the function Derived& operator = (const Base&) does not exist.

Instead, it called a constructor, even though the expression d = b is an assignment expression.

So why does this happen?

Answer 1

d = b can happen because b is converted to Derived.
The second constructor is used for automatic type conversion.
It’s like d = (Derived) b

Derived isa Base, but Base isn’ta Derived, so it has to be converted before assignment.