https://stackoverflow.com/questions/2337213/return-value-of-operator-overloading-in-c
I have a question about the return value of operator overloading in C++. Generally, I found two cases, one is return-by-value, and one is return-by-reference. So what's the underneath rule of that? Especially at the case when you can use the operator continuously, such as cout<<x<<y.
For example, when implementing a + operation "string + (string)". how would you return the return value, by ref or by val.
回答：
Some operators return by value, some by reference. In general, an operator whose result is a new value (such as +, -, etc) must return the new value by value, and an operator whose result is an existing value, but modified (such as <<, >>, +=, -=, etc), should return a reference to the modified value.
For example, cout is a std::ostream, and inserting data into the stream is a modifying operation, so to implement the << operator to insert into an ostream, the operator is defined like this:
std::ostream& operator<< (std::ostream& lhs, const MyType& rhs) { // Do whatever to put the contents of the rhs object into the lhs stream return lhs; }This way, when you have a compound statement like cout << x << y, the sub-expression cout << x is evaluated first, and then the expression [result of cout << x ] << y is evaluated. Since the operator << on x returns a reference to cout, the expression [result of cout << x ] << y is equivalent to cout << y, as expected.
Conversely, for "string + string", the result is a new string (both original strings are unchanged), so it must return by value (otherwise you would be returning a reference to a temporary(而且输入的也是const 引用对象，故只会在重载的函数中重新定义一个临时变量来存放结果), which is undefined behavior).
