Why should the assignment operator return a reference to the object?

Question

I'm doing some revision of my C++, and I'm dealing with operator overloading at the minute, specifically the "="(assignment) operator. I was looking online and came across multiple topics discussing it. In my own notes, I have all my examples taken down as something like

class Foo
{
    public:  
        int x;  
        int y;  
        void operator=(const Foo&);  
};  
void Foo::operator=(const Foo &rhs)
{
    x = rhs.x;  
    y = rhs.y;  
}

In all the references I found online, I noticed that the operator returns a reference to the source object. Why is the correct way to return a reference to the object as opposed to the nothing at all?

Answer 1

The usual form returns a reference to the target object to allow assignment chaining. Otherwise, it wouldn't be possible to do:

Foo a, b, c;
// ...
a = b = c;

Still, keep in mind that getting right the assigment operator is tougher than it might seem.

Answer 2

The return type doesn't matter when you're just performing a single assignment in a statement like this:

x = y;

It starts to matter when you do this:

if ((x = y)) {

... and really matters when you do this:

x = y = z;

That's why you return the current object: to allow chaining assignments with the correct associativity. It's a good general practice.