为什么我们在赋值运算符重载中使用引用返回而不是在加减运算中？

Question

As I read in books and in the web, in C++ we can overload the "plus" or "minus" operators with these prototypes (as member functions of a class Money ):

const Money operator +(const Money& m2) const;

const Money operator -(const Money& m2) const;

and for the assignment operator with:

const Money& operator =(const Money& m2);

Why use a reference to a Money object as a return value in the assignment operator overloading and not in the plus and minus operators?

Answer 1

Returning a reference from assignment allows chaining:

a = b = c;  // shorter than the equivalent "b = c; a = b;"

(This would also work (in most cases) if the operator returned a copy of the new value, but that's generally less efficient.)

We can't return a reference from arithmetic operations, since they produce a new value. The only (sensible) way to return a new value is to return it by value.

Returning a constant value, as your example does, prevents move semantics, so don't do that.

Answer 2

Because operator+ and operator- don't act on this object, but return a new object that is the summation (or subtraction) of this object from another.

operator= is different because it's actually assigning something to this object.

operator+= and operator-= would act on this object, and are a closer analog to operator= .

Answer 3

Consider what you are asking. You would want an expression, a + b , to return a reference to one of a or b, which would have the results of the expression. Thus you would modify one of a or b to be the sum of a and b. So you would want to redefine the semantics of the operator (+) to be the same as the operator (+=). And like @manuell said, you would thus allow (a + b) = c . The semantics you are suggesting are already offered by += and -=.

Answer 4

下面显示的链接有更好的解释我猜测C ++中运算符重载的返回值

Answer 5

I think its fine if you return by value in overloaded assignment operator , that is because of associativity of assignment operator. consider this:

int a = b =c = 3 ;

here associativity is as followed: (a=(b=(c=3)))

but consider iostream operation cout << x << y << z ;

here associativity is as followed: (((cout << x )<< y) << z) ;

you can see that x will be printed first , so if you return by value in overloading of << operator , return value will not be "lvalue" , while returning by refrence is a lvalue , so cascading of << operator can be achieve.

one more point , copy constructor will get called if you return by value. ( which is not the case with return by refrence)

Answer 6

Apart from efficiency during chained assignment (eg, a = b = c; ) as pointed out by the accepted answer, it is also a matter of convention. Please, refer Item 10 of Effective C++ by Scott Meyers. It is a convention followed by the built-in types and all the types in the standard library.