Re: Portability of comparing pointers to void

"Matthias Hofmann" <>
Mon, 29 Oct 2007 15:41:23 CST
"Jiri Palecek" <> schrieb im Newsbeitrag

Matthias Hofmann wrote:

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").
you have in mind would probably mean "return false" here.

With "return false", I end up with only a single element in the set: 2.

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.

If you created such a set, it would only consist of one element (in this
example, 2, AFAIK).

I just learned that my program crashes with 'a != b'. It seems like a strict
weak ordering cannot be implemented, only a total ordering. Otherwise, how
do you create a set of elements that are both distinct and incomparable?

have searched for a real life example for a strict weak ordering that is
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
you take your strict-weakly-ordered universe and partition it to classes
induced by the "incomparability" relation, you always get a linearly
ordered universe.

But ordering strings is a total ordering, isn't it? Because for strings, if
neither x < y nor y < x, the strings are equal. What about defining the
relation "<" as "is a descendant of" in a familiy tree? It satisfies the
condition that if x is incomparable with y and y is incomparable with z,
then x is incomparable with z. In other words, if x and y are siblings and y
and z are siblings, then x and z are siblings. It's also not a total
ordering because being siblings does not mean being equal.

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
know whether two elements are equal in order to determine whether the

No, the containers treat such two elements as equal, without calling

Yes, it uses operator< to find out wether two elements are equal. But that's
just my point: In a strict weak ordering, the relation "neither a < b nor b
< a" does not mean equality. Equality requires a total ordering, so how come
the standard requires only a strict weak ordering for associative

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.

Well, giving guarantees for user types is generally difficult. It is up to
the user to design their types to comply with the standard requirements.

Matthias Hofmann
Anvil-Soft, CEO - The Creators of Toilet Tycoon - Die Macher des Klomanagers

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