------- Comment #7 from pinskia at gcc dot gnu dot org 2007-04-22 18:18 -------
13.5.1/1 explains that:
@x is the same as operator@(x) or x.operator@() [depending on if x has a member
function for operator@ or not] .
So:
a = b++; is the exactly the same as:
a = b.operator++(0);
so function operator++ can do anything, it does not have to follow what is a
postfix incremement at all. 13.5/7 goes in a little detail at what is required
for basic types does not have to hold for overloaded operators.
So again this bug is invalid. The code is acting exactly what the C++ standard
says it should act and any change is going to make GCC a non C++ compiler.
>In these cases the current interpretation will by necessity involve a
> complete copy.
No, no, no, it does not this is one of my points I have been trying to explain,
12.8/15 explains sometimes temporaries can be avioded and this case it can.
For any:
a = b.function();
where function could be an operator call (invoke implictly via the operator
too) and the function returns a large data type so the copy constructor is not
called and there is no extra variable on the stack. So I don't see your point
with postfix being special really, as there is nothing special about those
operators.
--
pinskia at gcc dot gnu dot org changed:
What |Removed |Added
----------------------------------------------------------------------------
Status|UNCONFIRMED |RESOLVED
Resolution| |INVALID
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=31652
