I am reading the C++ Primer, in the overloaded operation chapter, the author gave an example:
// member binary operator: left-hand operand bound to implicit this pointer
Sales_item& Sales_item::operator+=(const Sales_item&);
// nonmember binary operator: must declare a parameter for each operand
Sales_item operator+(const Sales_item&, const Sales_item&);
then, the author explained:
This difference matches the return types of these operators when applied to arithmetic types: Addition yields an rvalue and compound assignment returns a reference to the left-hand operand.
I'm not quite sure about "compound assignment returns a reference to the left-hand operand". Can anyone elaborate on that, and relevant things, please?
It means that you can do something like the following
a = 1;
(a += 1) += 1;
and the result will be a == 3. This is because the left most call to += modifies a and then returns a reference to it. Then the next += operates on the reference to a and again adds a number to it.
On the other hand, the normal + operator returns a copy of the result, not a reference to one of the arguments. So this means that an expression such as a + a = 3; is illegal.
