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

[Eric Lafontaine]

Eric Lafontaine added the comment:

Hi all,

Here are the test I've made to understand the behavior :

class Foo_42(object):
    def __contains__(self,item): return 42 
    
class Foo_neg(object):
    def __contains__(self,item): return -42 
    
class Foo_None(object):
    def __contains__(self,item): return  
    
class Foo_false(object):
    def __contains__(self,item): return False 
    
class Foo_true(object):
    def __contains__(self,item): return True

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

which output this :
3 in foo:False
foo.__contains__(3)False
3 in foo:False
foo.__contains__(3)None
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 as long as __contains__ return False or None, the 'in' operator will be 
False.  Otherwise true.

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Python tracker <rep...@bugs.python.org>
<http://bugs.python.org/issue16011>
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