Thread: Copy Construtor

Run the code with this function: (notice the ref)

Code:

A f(A& x)
 {
  return x;
 }

Can you tell the difference?

You did that:

Code:

A b= f(a);

So, when f(a) was called, what happened to a? Moreover, look at the definition of the function named "f()". The argument is neither a reference nor a pointer. So, we have to copy the a from main to the argument of your function. So, copy constructor was called.
The second call comes from the assign operator, but I think you had figured that already.

When you use the function with the reference, the copy constructor get's called once, because we pass a reference in the function to the already constructed object a.

But, imagine that when the function gets called, the object comes from main and gets assigned to the argument of the function, thus we can imagine it as an assignment in that case, which of course calls the copy constructor and not the real constructor.

If I understand correct, you mean this:

Code:

int main()
{
   A a;
   A c;
   c = a;
   return 0;
}

In this case, no copy constructor is called, because both 'a' and 'c' objects are already constructed.

Originally Posted by std10093

The second call comes from the assign operator, but I think you had figured that already.

No, first call comes from copying a to the function.
Second call comes from returning a local variable by value.
Remove the assignment and it will still be the same.

Originally Posted by nickman

So it means, when ever we pass object as value and not reference, a new object is created which called copy constructor.

Correct.

But I am curious to know why only copy constructor is called and not the normal constructor?

Think about it: a copy constructor is also a constructor which constructs an object by copying over the state from another object.
A default constructor is a constructor that constructs an object with an empty state.

Is it by design that only copy constructor got called when object pass as value to function?

Yes, due to the above.

Originally Posted by nickman

Thanks Again.

Again I think that if I do this ( as you are saying its assignment):

Then again copy constructor should called, but it doesn;t.

That is assignment which means that the copy constructor should not be called. Instead, the copy assignment operator is called.
You should separate

A a;
A b(a); // Or A b = a;

and

A a, b;
a = b;

The former is copy construction and the later is copy assignment.

Originally Posted by Elysia

No, first call comes from copying a to the function.
Second call comes from returning a local variable by value.
Remove the assignment and it will still be the same.

Oh Elysia is correct. How in the world did I miss that? Sorry.

If you run this code, you will see what Elysia says:

Code:

A& f(A& x)
 {
  return x;
 }
 
int A:: count = 0;
 
int main()
{
   A a;
   f(a);
   return 0;
}