Unexpected behaviour from the 'in' operator

[Ronald Oussoren]

On Monday, Feb 24, 2003, at 16:06 Europe/Amsterdam, sik0fewl wrote:

> Blair Hall wrote:
>> I have just noticed that the 'in' operator for lists appears to
>> use the __eq__ method of a class. This does not
>> provide the behaviour I expected when implementing
>> an ad hoc numeric class type.
>> Here is some code:
>> class A(object):
>>     def __init__(self,x):
>>         self.x = x
>>     def __eq__(self,other):
>>         print 'eq'
>>         return self.x == other
>> #=============================
>> if(__name__ == '__main__'):
>>     print
>>     x = A(1)
>>     y = A(1)
>>     lst = [x]
>>     print y in lst        # prints 1
>> The 'y in lst' expression evaluates to true here, because
>> both x and y are equal to the same value. However,
>> x and y are not the same object!  It seems to me that
>> the comparison should use something like the id() function
>> instead of __eq__.
> I don't know. It works the way I would've expected it to work. 
> Because, like you said, x == y and x is in list, therefore y is in 
> list.

And if you use the 'is' operator instead of __eq__ you'll get very 
confusing behaviour:
	id(1000) != id(500+500)	 # For large enough integers :-)
	id("foo"+"bar") != id("foobar")

Ronald