Question

Look at the following code:

#include <iostream>
using namespace std;

class Widet{
public:
    Widet(int val = 0):value(val)
    {

    }

    Widet& operator=(Widet &rhs)
    {
        value = rhs.value;
        return *this;
    }
    int getValue()
    {
        return value;
    }
private:
    int value;
};

int main()
{
    Widet obj1(1);
    Widet obj2(2);
    Widet obj3(0);
    (obj3 = obj2) = obj1;
    cout << "obj3 = " << obj3.getValue() << endl;
}

The code runs successfully and the output is (using VS2008):

When I let the operator= return a value instead of reference:

Widet operator=(Widet &rhs)
{
    value = rhs.value;
    return *this;
}

It also runs successfully and the output is :

My question is :Why the second code runs well?Should not we get a error?

Why it is a good habit to return reference to *this instead of *this?

Answer 1

Why the second code runs well?Should not we get a error?

Because it’s perfectly valid code. It returns a temporary copy of the object, and you’re allowed to call member functions (including operator=()) on temporary objects, so there is no error.

You would get an error if the object were uncopyable.

Why it is a good habit to return reference to *this instead of *this?

Because not all objects are copyable, and some objects are expensive to copy. You can take a reference to any object, and references are always cheap to pass around.