This is clearly not working properly.
Why is this happening?
>>> d = {0 : "a", 1 : "b", 2 : "c", 3 : "d", True: 99}
>>> print d
{0: 'a', 1: 99, 2: 'c', 3: 'd'}
>>>
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notMarc
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2Because `True == 1` is true. – Martijn Pieters Feb 22 '16 at 22:15
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2This is a duplicate, an often-encountered trap. You can't put in `False` either, because `False == 0`. – Martijn Pieters Feb 22 '16 at 22:15
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Or rather, whichever type is added as a key first, wins; `{True: 'boolean', 1: 'integer'}` will leave you with `{True: 'integer'}`.. – Martijn Pieters Feb 22 '16 at 22:18
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@DeveloperXY - Uh.. it prints `True`. Did you mean `print int(True)`? – BingsF Feb 22 '16 at 22:28
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@BingsF In deed, thanks for the heads up – Mohammed Aouf Zouag Feb 22 '16 at 22:35
1 Answers
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True == 1. There are reasons to disagree with that design decision, but that's how things are in Python. (Similarly, False == 0.) You can't have two equal keys in a dict, so you can't have both True and 1 as keys in the same dict.
user2357112
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