Returning by reference

[logan]

class Foo
{
	int data;

public:
	Foo (int n):data (n)
	{
	}

	Foo ():data (0)
	{		
	}

	Foo &operator= (int n)
	{
		data = n;
		return *this;
	}

	Foo &operator= (const Foo& foo)
	{
		data = foo.data;
		return *this;
	}

	void display ()
	{
		cout << "data = " << data << endl;
	}
};

Why in the above code the operator= returns by reference? If I return it by value (Foo operator= ()) it seems to work fine. Do we return by refernce because it is faster (only the address of the object will be returned instead of making a copy of the object and then returning it, please correct me if I'm wrong )?

Thanks.

[AlbertGM]

To be able to do something like:

a = b = c;

Albert.

[GNiewerth]

Hi Albert,

your assignment can be done with return by value operators, too. References are returned to modify the return value in a later expression:

// assign a + b to c, multiply c by d and store the result in c.
(c = a + b) *= d;

This cannot be done by returning by value, because it changes a temporary object only, leaving c untouched (after it has been assigned a+b), so you have to return by reference.

Regards,
Guido

[AlbertGM]

Hi Guido,

Really? So, I was wrong!!! I thought the reference was necessary to make the last assignation

a=b=c;

Thanks.

Albert.

[logan]

Guys, thank you for the responses [].

[Zaccheus]

It's also for efficiency.

	Foo &operator= (const Foo& foo)
	{
		data = foo.data;
		return *this;
	}

[Krishnaa]

Hi Guido,

Really? So, I was wrong!!! I thought the reference was necessary to make the last assignation

a=b=c;

Thanks.

Albert.

Partly true, not wrong.