Why is the copy constructor not called?

Question

class MyClass
{
public:
  ~MyClass() {}
  MyClass():x(0), y(0){} //default constructor
  MyClass(int X, int Y):x(X), y(Y){} //user-defined constructor
  MyClass(const MyClass& tempObj):x(tempObj.x), y(tempObj.y){} //copy constructor

private:
  int x; int y;
};

int main()
{
  MyClass MyObj(MyClass(1, 2)); //user-defined constructor was called.
  MyClass MyObj2(MyObj); //copy constructor was called.
}

In the first case, when MyClass(1, 2) calls the user-defined constructor and returns an object, I was expecting MyObj to call the copy constructor. Why it doesn't need to call the copy constructor for the second instance of MyClass?

Answer 1

Whenever a temporary object is created for the sole purpose of being copied and subsequently destroyed, the compiler is allowed to remove the temporary object entirely and construct the result directly in the recipient (i.e. directly in the object that is supposed to receive the copy). In your case

MyClass MyObj(MyClass(1, 2));

can be transformed into

MyClass MyObj(1, 2);

even if the copy constructor has side-effects.

This process is called elision of copy operation. It is described in 12.8/15 in the language standard.