28 Apr
2005
28 Apr
'05
4:23 p.m.
?? The return type of the generated operators is perfectly well visible in the documentation.
The documentation (http://www.boost.org/libs/utility/operators.htm) says that the operators return "convertible to T".
I think that the user will expect that the prototype of function match exactly those in table... Also, as far as I understand convertible to T is a requirement for the implementation... and in fact it does says nothing as the requirement are (generally) in the form T temp(*this); temp op= u; // +=, -=, ... and is complete by itself. Philippe