[Python-Dev] Rich comparison confusion

[Guido van Rossum]

> I'm a bit confused about Guido's rich comparison stuff.  In the description
> he states that __le__ and __ge__ are inverses as are __lt__ and __gt__.

Yes.  By this I mean that A<B and B>A are interchangeable, ditto for
A<=B and B>=A.  Also A==B interchanges for B==A, and A!=B for B!=A.

> From a boolean standpoint this just can't be so.  Guido mentions partial
> orderings, but I'm still confused.  Consider this example: Objects of type A
> implement rich comparisons.  Objects of type B don't.  If my code looks like
> 
>     a = A()
>     b = B()
>     ...
>     if b < a:
>         ...
> 
> My interpretation of the rich comparison stuff is that either
> 
>     1. Since b doesn't implement rich comparisons, the interpreter falls
>        back to old fashioned comparisons which may or may not allow the
>        comparison of B objects and A objects.
> 
>     or
> 
>     2. The sense of the inequality is switched (a > b) and the rich
>        comparison code in A's implementation is called.

It's case 2.

> That's my reading of it.  It has to be wrong.  The inverse comparison should
> be a >= b, not a > b, but the described pairing of comparison functions
> would imply otherwise.

We're trying very hard *not* to make any connections between a<b and
a>=b.  You've learned in grade school that these are each other's
Boolean inverse (a<b is true iff a>=b is false).  However, for partial
orderings this may not be true: for unordered a and b, none of a<b,
a<=b, a>b, a>=b, a==b may be true.

On the other hand, even for partially ordered types, a<b and b>a
(note: swapped arguments *and* swapped sense of comparison) always
give the same outcome!

> I'm sure I'm missing something obvious or revealing some fundamental failure
> of my grade school education.  Please explain...

I think what threw you off was the ambiguity of "inverse".  This means
Boolean negation.  I'm not relying on Boolean negation here -- I'm
relying on the more fundamental property that a<b and b>a have the
same outcome.

--Guido van Rossum (home page: http://www.python.org/~guido/)