Question

I stumbled across a question that I never thought about before.
Here it is:
each object’s (listed in the initialization list) “constructor” will be triggered.

class B
{
    public:
        B() { cout<<"B Con\n";}
        B(const B &b) { cout<<"B Copy Con\n";}
};

class A
{
    public:
        A(B &b):_m(b) { cout<<"A Con\n";}
        A(const A &a):_m(a._m) { cout<<"A Copy Con\n";}
    private:
        B _m;
}

main()
{
    B b;
    A a(b);
}

then I got the output as follows:

B Con
B Copy Con
A Con

According to the output, I think, ‘A a(b)’ triggered B’s copy constructor.
If I got right, then that means ‘A(B &b):_m(b)’ triggers B’s copy constructor.
Why not constructor but copy-constructor?

Answer 1

The reason is when you call

_m( whatever )

then the copy constructor

B(const B &b)

is the only one that could match the parameter list. You pass it one parameter and that parameter is of type class B.

Copy constructor is not something super special – it is just a parameterized constructor that will be invoked via the initialization list once the parameter list matches.