Re: using operator++() without its side-effect
On Apr 9, 1:59 am, Daniel Kr?gler <daniel.krueg...@googlemail.com>
wrote:
On 8 Apr., 19:18, itaj sherman <itajsher...@gmail.com> wrote:
On Apr 8, 12:32 pm, Ulrich Eckhardt <eckha...@satorlaser.com> wrote:
Here's one cake for you then, to both eat and keep:
template<typename T>
T copy(T const& t)
{ return t; }
iterator i0 = ...;
iterator i1 = ++copy(i0);
:)
Uli
I'm not sure I get this: how does the temporary returned by copy bind
to a reference parameter of operator++. Is it some rvalue-reference
version of operator++? Can it be done in c++03?
There is no reference binding intended. I think
that Uli just wants to demonstrate that above
code is not guaranteed to be well-formed: For
built-in pointers an lvalue is required for the
preincrement operator, but copy returns an rvalue.
HTH & Greetings from Bremen,
Daniel Kr?gler
So, the code doesn't compile unless c++0X with an ::operator++
( iterator&& ) overload, right?
1) As I said: the parameter of operator++ is non-const reference and
cannont bind to the temporary returned by copy.
2) As you said: operator++ expects lvalue parameter and cannot accept
rvalue returned by copy.
I thought there isn't a difference between 1 and 2, is there?
itaj
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