Re: Portability of comparing pointers to void

From:
Greg Herlihy <greghe@mac.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Mon, 29 Oct 2007 14:57:43 CST
Message-ID:
<1193650107.835281.196600@t8g2000prg.googlegroups.com>
On Oct 28, 11:02 am, "Matthias Hofmann" <hofm...@anvil-soft.com>
wrote:

"Matthias Hofmann" <hofm...@anvil-soft.com> schrieb im Newsbeitragnews:5ofg50Fmfjn1U1@mid.individual.net...

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, the above functor provides no ordering at all between a and b. For
a and b to be ordered, the result of evaluating a < b and b < a must
match one of the three cases listed below:

           a < b b < a
     1. true false
     2. false true
     3. false false

The above functor (when a != b) reports:

           a < b b < a
     1. true true

a truth table which clearly makes it impossible to decide which of the
two objects ("a" or "b") should be ordered before the other.

(I
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?)


Just define a class that implements less-than comparisons between
class objects but does not implement equality comparisons between
class objects.

The existence of a strict weak ordering for a set of objects, means
that it is possible to tell whether one object precedes another (or
whether the two objects have the same precedence). But a strict weak
ordering does not guarantee that it is possible to tell whether two
objects are "equivalent".

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?


Because pointers can be compared for equality (with the equality
operator: ==). So a strict weak ordering for pointers becomes a total
ordering because an equality comparison is defined. (Note that a
strict weak ordering also imposes a total ordering, if

      (not(a < b) and not(b < a))

defines an equivalence relation between a and b.)

Shouldn't the standard require a total ordering for
std::less for *any* type, not just for pointers?


Then a type would have to support equality comparisons in addition to
relational comparisons, in order to qualify as a type parameter to
std::less<>. Yet equality comparisons may not make much sense for
certains classes - nor is it clear what imposing this additional
requirement would accomplish.

Greg

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