Re: Portability of comparing pointers to void
Matthias Hofmann wrote:
"Matthias Hofmann" <hofmann@anvil-soft.com> schrieb im Newsbeitrag
news:5ofg50Fmfjn1U1@mid.individual.net...
"Hyman Rosen" <hyrosen@mail.com> schrieb im Newsbeitrag
news:eobUi.22974$Qj3.7738@trndny01...
Matthias Hofmann wrote:
Meanwhile I did some further research on partial ordering, strict weak
ordering and total ordering and things are a little clearer for me now,
but there are still two things that keep me wondering:
1.) As far as I understand strict weak ordering, two elements of a set may
be incomparable as long as this incomparability relation is transitive. If
this is true, then the following should be fine?
#include <set>
template <class T> struct strict_weak_less;
template <> struct strict_weak_less<int>
{
bool operator()( int a, int b )
{
// Define any two elements
// to be incomparable unless
// they are equal.
return a != b;
No. What you have written is not a strict weak ordering, because it lacks
antisymmetry (you say "a is less then b whenever they are not equal"). What
you have in mind would probably mean "return false" here.
}
};
int main()
{
std::set<int, strict_weak_less<int> > intset;
intset.insert( 2 );
intset.insert( 4 );
return 0;
}
This creates a set where all elements are incomparable with each other. (I
If you created such a set, it would only consist of one element (in this
example, 2, AFAIK).
have searched for a real life example for a strict weak ordering that is
not
a total ordering, but I have not found any. Does anyone know an example?)
Sure. For example, ordering of strings or sets by their sizes. Note, that if
you take your strict-weakly-ordered universe and partition it to classes
induced by the "incomparability" relation, you always get a linearly
ordered universe.
2.) Why does the standard require a strict weak ordering for associative
containers, but guarantee a total ordering for std::less<T*> in
20.3.3/8? In
a strict weak ordering, the relation "neither a < b nor b < a" is
transitive, but does not mean equality. Incomparability only means
equality in a total ordering, but doesn't an associative container need to
know whether two elements are equal in order to determine whether the
No, the containers treat such two elements as equal, without calling
operator==.
element already exists? Shouldn't the standard require a total ordering
for std::less for *any* type, not just for pointers?
It would be impossible to guarantee that for user types.
Regards
Jiri Palecek
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