[c++]: problem about initiation list in constructor

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.

Answer 2

Because you're telling the compiler to initialize _m with b, how would that not call the copy constructor?

Answer 3

The answer lies in the A(B &b):_m(b) You are instantiating B _m with the copy constructor.

If instead you did A(B &b):_m() it would use the default constructor.

Answer 4

A(B &b):_m(b) { cout<<"A Con\n";}

Here _m(b) causes invocation of B(const B&) which is B's copy-constructor. That is why, it first prints B Copy Con when initializing _m, then it enters into A's constructor body, and prints A Con. That explains it all.