Re: Returning Refs
<EvilOldGit@googlemail.com> wrote in message
news:1175349463.536538.36070@y80g2000hsf.googlegroups.com...
const Thing &operator++(int)
{
Thing temp = *this;
operator++();
return temp;
}
Is this code robust ?
No. You are returning a reference to a temporary item. temp gets destroyed
when this mothod returns and so your'e returning a reference to something
that no longer exists.
Just return Thing, not Thing&
I get a compiler warning about returning a reference to a a local,
which I guess is because the compiler doesn't tell the difference
between this overload any any other function.
That pretty much makes no sense to me whatsoever. The compiler is warning
you that you are returning a reference to a local, because you are returning
a reference to a local. Your compiler is not lying to you, you are lying to
yourself.
Thing x,y;
x = y++;
Is OK because it's thrown away, and it's a copy of non-incremented y
that gets assigned.
No, that's not okay if you are reutrning a reference to x, because it is
also a local that will get destoryed when the method is finished executing.
y++ = x;
Does things to my head, but can't happen because I'm returning a
const.
const_casting<> doesn't
But could a (creatively stupid) user of my code do some harm ?
Any user of your code would do some harm. Your code is wrong.