>>> l = [False, True]
>>> l.index(True)
1
>>> l.index(False)
0

good, but:

>>> l = ['a', '', {}, 2.7, 1, 0, False, True]
>>> l.index(True)
4
>>> l.index(False)
5

Apparently, True and False get converted to int in comparisons to ints.
I would expect items to be compared either by:

a) object identity or
b) __eq__

but not this inconsistently.

Best,
Wolfgang

It's using __eq__:
>>> l = [3, 1.0, 5, True, 7, 1]
>>> l.index(1)
1
>>> l.index(True)
1
>>> l.index(1.0)
1
>>> True == 1 == 1.0
True

No, it's not that simple and I don't think this should be closed:

In my example:
>>> l = ['a', '', {}, 2.7, 1, 0, False, True]
>>> l.index(True)
4
>>> l.index(False)
5

if using __eq__ consistently, you'd expect the first call to return 0 and the second 1 (since the empty string evaluates to False).

'a' evaluates to true, but it's not equal to True:
  >>> bool('a')
  True
  >>> 'a' == True
  False
but 1 and True are equal (for historical reasons):
  >>> 1 == True
  True

Similarly '' evaluates to false, but it's not equal to False:
  >>> bool('')
  False
  >>> '' == False
  False
whereas 0 is equal to False:
  >>> 0 == False
  True

Right, forgot about that.

The consequence for the example is still far from satisfying, I think, but you can't change it without breaking compatibility then.

Thanks for the quick reply,
Wolfgang