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?
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.
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=.
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 -=.
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)
The link shown below has better explanation I guess return value of operator overloading in C++
来源：https://stackoverflow.com/questions/21485366/why-we-use-reference-return-in-assignment-operator-overloading-and-not-at-plus-m