how to overload operator "< <" (a < x < b)?

[Diez B. Roggisch]

alex23 schrieb:
> On Aug 7, 10:50 pm, Benjamin Kaplan <benjamin.kap... at case.edu> wrote:
>> That isn't an operator at all. Python does not support compound
>> comparisons like that. You have to do "a > b and b > c".
> 
> You know, it costs nothing to open up a python interpreter and check
> your certainty:
> 
>>>> x = 10
>>>> 1 < x < 20
> True
> 
> This is a _very_ common pattern.
> 
>>>> class X(object):
> ...   def __lt__(self, other):
> ...     print 'in lt'
> ...     return True
> ...   def __gt__(self, other):
> ...     print 'in gt'
> ...     return True
> ...
>>>> x = X()
>>>> 1 < x < 20
> in gt
> in lt
> True
>>>> 20 < x < 1
> in gt
> in lt
> True
> 
> dmitrey: Diez' advice was the best you received.

Not really. I didn't get the chaining, and Peter is right that for that 
there is no real overloading.

Diez