Operator =：返回参考值与值

Question

I was going through this and am a bit confused. Suppose that I declare a class as:

class cls
{
public:
    int x;
    cls(int _x):x(_x){}
    cls& operator=(cls& ob)
    {
        x = ob.x;
        return *this;
    }
};

And then create 2 objects and perform copy operation and then print the addresses of both the variables before and after the assignment operator is overloaded as:

cls o1 = 7;
cls o2 = cls(8);
cout<<&o1<<endl;    //0330F880
cout<<&o2<<endl;    //0330F874
o1 = o2;
cout<<&o1<<endl;    //0330F880
cout<<&o2<<endl;    //0330F874

Both the address group is same; this is understood as the assignment operator returns by reference.

But I notice that the same address group values are returned if I define my assignment operator to return by value.

In the link referred above, It is answered that a copy of the object will be returned if returned by value. Then why is it returning the same address values. Shouldn't they be different. Please help clear my concepts.

Answer 1

The return value is only relevant if you do something with it. For example:

(o1 = o2).do_something();

Or equivalently:

(o1.operator=(o2)).do_something();

The do_something() method will run on the object returned - in your case the original instance of o1 since it was returning a reference. However, if you changed your code to return a value instead, then do_something() would be running on a copy of o1 .

If you had a third object cls* o_ptr; and did the following:

cls o1 = 7;
cls o2 = cls(8);
cls* o_ptr = &(o1=o2);

If you displayed o_ptr you'd see it was the same as &o1 if you return a reference, but different if you returned a value.

Answer 2

我认为，这个地址是一样的，因为你只是覆盖了结构的内部内容，而不是它在内存中的位置。