C ++ operator =返回* this的引用

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.

Answer 2

通常，您返回一个引用，以便可以使用=运算符作为l值。

Answer 3

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

It runs because nonconst member functions can be called on (nonconst) class rvalues as well. The second version of operator= returns a nonconst class rvalue, so in effect, you assign to the temporary, leaving the previous value in the obj3 variable.

Therefore, there is no error.

Answer 4

When you don't return a reference, (obj3 = obj2) gives a temporary copy of obj3 . The copy obtains the value from obj1 and is deleted, while obje3 is never affected by the second assignment.

Answer 5

In the second example you create a temporary (a copy of Obj3, returned by operator=) and assign Obj1 to it. Then it immediately gets destructed. Obj3 remains the result of first assignment - Obj3 = Obj2 .

Answer 6

Returning a reference from operator=() enables expressions like:

a=b=c;

Returning a value may be excessive when you don't need it. It can cause extra copy-constructor/destructor calls. Otherwise, returning a value is perfectly valid C++. People, please correct me if I'm wrong, but I think returning by value is not that big of an issue in C++11 because of move semantics.

Answer 7

(obj3 = obj2)

can be considered as obj3operator=(obj2) //hypothetically.
Since you have passed obj2 as parameter, your operator overload will copy the obj2.value into obj3.value.

(obj3 = obj2) = obj1;
After the return of from operator= , obj3 (*this,temporary copy) will be returned.
So the equivalent code becomes obj3=obj1 which will again invoke operator= of obj3 and will reset value of obj3.value to obj1.value ie 1 .