return value of operator overloading in C++

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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, which is undefined behavior).

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