[docs] [issue16011] "in" should be consistent with return value of __contains__

[Eric Lafontaine]

Eric Lafontaine added the comment:

(first time trying to reply through email)

thanks for the example and you are right :
class Foo_emptylist(object):
    def __contains__(self,item): return []

class Foo_emptydict(object):
    def __contains__(self,item): return {}

class Foo_emptystring(object):
    def __contains__(self,item): return ''

for foo in
[Foo_false(),Foo_None(),Foo_emptylist(),Foo_emptydict(),Foo_emptystring(),Foo_neg(),Foo_true(),Foo_42()]:
    print("3 in foo:" + str(3 in foo))
    print("foo.__contains__(3)" + str(foo.__contains__(3)))

3 in foo:False
foo.__contains__(3)False
3 in foo:False
foo.__contains__(3)None
3 in foo:False
foo.__contains__(3)[]
3 in foo:False
foo.__contains__(3){}
3 in foo:False
foo.__contains__(3)
3 in foo:True
foo.__contains__(3)-42
3 in foo:True
foo.__contains__(3)True
3 in foo:True
foo.__contains__(3)42

So the proposition should be this then?
For user-defined classes which define a __contains__() method, the in
operator will apply bool() on the __contains__() method.  In other words,  "x
in y" is equivalent to "bool(y.__contains__(x))" and will return False if
bool(y.__contains__(x)) is equivalent to false.

Éric Lafontaine |  Membre du Projet VUE, Groupe Contrôle
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2017-02-07 21:10 GMT-05:00 R. David Murray <report at bugs.python.org>:

>
> R. David Murray added the comment:
>
> >>> bool(())
> False
> >>> bool([])
> False
> >>> bool('')
> False
>
> What you want to say is that 'in' coerces the result returned by
> __contains__ to a boolean value.
>
> ----------
>
> _______________________________________
> Python tracker <report at bugs.python.org>
> <http://bugs.python.org/issue16011>
> _______________________________________
>

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Python tracker <report at bugs.python.org>
<http://bugs.python.org/issue16011>
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