Re: std::set<> and predicates
On 5 oct, 12:35, Rune Allnor <all...@tele.ntnu.no> wrote:
On 5 Okt, 09:46, James Kanze <james.ka...@gmail.com> wrote:
On Oct 5, 4:55 am, Rune Allnor <all...@tele.ntnu.no> wrote:
I am a bit confused about std::set<> and associated
predicates. I have a std::set in my application where the
predicate works as expected when tested in isolation, but
where the items inserted in the set don't match the predicate,
that is, the test
class predicate : public std::less {/*...*/};
predicate p;
std::set<size_t> s(p);
// Insert elements
std::set<size_t>::iterator i = s.begin();
std::set<size_t>::iterator j = i; j++;
bool b = p(*i,*j); // Should be true
Knowing nothing about p, it's impossible to say much. But two
things are obvious: as declared above, your set doesn't use
predicate as an ordering function, it uses std::less<size_t>;
Why is that?
According to Josuttis' table 6.20, the constructor std::set<>::set()
with the predicate as argument should use the predicate as ordering
criterion.
set<> constructor keeps a copy of the Comparator. In your case, you
have slicing at the less<size_t> base of predicate.
I could not find the related section in the standard. I will post it
if/when I find it.
and if it did use predicate, the above expression could never be
true.
As I understand std::set and its predicates, the ordering of the
elements in the set is such that
less(*i,*j) == true
if i and j are iterators and initailized as above. If you are
right, I would have missed something very fundamental. Could you
elaborate, please?
Also, even the extract of your code is illegal, since you can't
derive from std::less (which is a template, not a class).
Actually, in my code the inheritance is
class p : public std::less<size_t> {
};
So how should this be done?
A way to do it is:
class predicate{
bool operator(size_t lhs,size_t rhs)const
{ /* ... */ }
};
std::set<size_t,predicate> p;
--
Michael