Returning by reference

Code:

---

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 :o)?

Thanks.

To be able to do something like:

Code:

---

a = b = c;

---

Albert.

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:

Code:

---

// 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

Hi Guido,

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

Code:

---

a=b=c;

---

Thanks.

Albert.

Guys, thank you for the responses [:)].

It's also for efficiency.

Code:

---

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

---

Quote:

---

Originally Posted by AlbertGM

Hi Guido,

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

Code:

---

a=b=c;

---

Thanks.

Albert.

---

Partly true, not wrong. :)