Re: Clean ways to count remove_if() removals?
On 24 Jun., 10:15, Michael Doubez <michael.dou...@free.fr> wrote:
On 24 juin, 09:45, SG <s.gesem...@gmail.com> wrote:
If remove_if returned a copy of the predicate you could have written
p = list.remove_if(p);
Which is not a big help since AFAIK there is no guarantee that the
same predicate instance is called on all elements or even that it is
called only once.
In addition to that, there is not much sense in returning a predicate,
it is not like it is supposed to hold a value.
Right.
That's how it works fo std::for_each.
But for_each() takes a functor, not a predicate and provides the
guarantee that it returns the function object after it has been
applied to each element (and local state are allowed).
Yeah, I know. If you check out the requirements of the concept
Predicate you'll note only two restrictions: A predicate returns
something that is implicitly convertible to bool and it is not allowed
to modify its arguments. You still can have "state" (i.e. pointer).
In C++0x you will be able to write
list.remove_if(ref(p));
I just forgot about lambdas:
int count = 0;
list.remove_if([&count](my_type const& x)->bool{...});
reference_wrapper<T> has an overloaded function call operator which
perfectly forwards parameters. I don't know whether Boost.Ref has any
magic in it to do something similar.
Which doesn't help either.
Please elaborate. I don't have the C++03 standard documents available
but the C++0x draft explicitly mentions
Complexity: Exactly distance(begin(), end()) applications of the
corresponding predicate.
which seems enough a guarantee to me to make the ref(p) or lambda
version work. It doesn't even matter if the function call operator is
applied on the /same/ predicate object.
Cheers!
SG