classes

[moo2401]

#include<iostream>

class A
{
public :
int*x;
};

class B
{
public:
A _a;

};

int main()
{
B b1;
b1._a.x = new int(1);
B b2(b1);
*b2._a.x = 2;
std::cout<< b2._a.x << std::endl;
std::cout<< b1._a.x << std::endl;
system("pause");
}





what does B b2(b1); mean ?

[Peter87]

It means the same as B b2 = b1;.
b2 is created as a copy of b1.

[moo2401]

so when using operator () its the same as using =
that is new and thanks peter for the help

[Enoizat]

so when using operator () its the same as using =

I’m afraid that’s not accurate.
In C++ you can overload operator(), and the expression “operator()” usually refers to that procedure. But here you’re getting advantage of implicitly-defined copy constructor
http://en.cppreference.com/w/cpp/language/copy_constructor
and implicitly-defined copy assignment operator
http://en.cppreference.com/w/cpp/language/copy_assignment

Note: in your example both of them might disappoint you, since they copy the pointer, not what is pointed to. Therefore, after being copy-constructed, both “x” member of b1 and b2 point to the same memory area. You normally don’t want that.