A Question on Copy Constructor

The copy constructor isn't called. Why?

Here is the code.

Code:

---

#include<iostream>

using namespace std;

class Circle
{
private:
      int x,y;
      int radius;
public:
      Circle();
      Circle(const Circle &);
};

Circle getCircle();

int main()
{
    Circle c;
    c = getCircle();
   
    cin.get();
    return 0;
}

Circle::Circle()
{
      x = y = radius = 30;
}

Circle::Circle(const Circle &c)
{
      cout << "copy constuctor!" << endl;
      x=c.x;
      y=c.y;
      radius = c.radius;
}

Circle getCircle()
{
      Circle c;
      return c;
}

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When getCircle() returns, it should call the copy constructor and generate a temporary object, right?
But it comes out that the copy constructor is not call cause I cannot find `copy constuctor!' in the output.

Your copy constructor should be called with that code.

Quote:

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Originally Posted by Antigloss

When getCircle() returns, it should call the copy constructor and generate a temporary object, right?

---

Couldn't default ctor + assignment operator be the culprit?

[C++ newb still reading this.]

Quote:

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Couldn't default ctor + assignment operator be the culprit?

---

I think that is it.

Antigloss, why not just add print statements to the default constructor and copy assignment operator, just like what you have done for the copy constructor?

The C++ standard allows the compiler to eliminate temporary objects if the only way of detecting the existance of those temporaries is by tracking calls to constructors or destructors.

Logically, the getCircle() function would create a temporary copy of c which would be returned to the caller. However, as the only way of detecting that temporary would be by tracking constructor and destructor calls ...... In this particular case, the compiler has implemented what is known as the "Return Value Optimisation".

Thanks grumpy and also the other guys.
Thank you very much!