Why constructor isn't called?

Question

Why constructor of class A isn't called when object of this class is passed as an argument to function taking the argument by value?

class A
{
public:
    A()
    {
        cout << "A\n";
    }
};

void f_n(A val)
{

}

int main()
{
    A a;
    f_n(a);
    return 0;
}

Because the copy constructor is called instead. - See https://ideone.com/vmI74w — Ed Heal, Jun 18 '16 at 09:00

score 0 · Accepted Answer · answered Jun 18 '16 at 09:09

This is because copy constructor is used in this context. To check it add explicit implementation:

A(A const&)
{
    std::cout << "A const&\n";
}

You have not defined one but compiler synthesizes copy constructor for you. But there are cases when it is not generated.

1 Answers1