Why does string > int evaluate to True?
python, python-2.7
Solution
Source: How does Python compare string and int?, which in turn quotes the CPython manual:
CPython implementation detail: Objects of different types except numbers are ordered by their type names; objects of the same types that don’t support proper comparison are ordered by their address.
From the SO answer:
When you order two incompatible types where neither is numeric, they are ordered by the alphabetical order of their typenames:
>>> [1, 2] > 'foo' # 'list' < 'str'
False
>>> (1, 2) > 'foo' # 'tuple' > 'str'
True
>>> class Foo(object): pass
>>> class Bar(object): pass
>>> Bar() < Foo()
True
...so, it's because 's' comes after 'i' in the alphabet! Luckily, though, this slightly odd behavior has been "fixed" in the implementation of Python 3.x:
In Python 3.x the behaviour has been changed so that attempting to order an integer and a string will raise an error:
Seems to follow the principle of least astonishment a little better now.
Problem
How come a check of string > int evaluates to True? ``` >>> strver = "1" >>> ver = 1 >>> strver > ver True >>> strVer2 = "whaat" >>> strVer2 > ver True ``` Did some more experimenting: ``` >>> ver3 = 0 >>> strVer2 > ver3 True ``` I think there should be an error when trying to compare but it seems like nothing is built to handle such an error, or `assert` should be used but that can be dangerous if python code is being run with -O flag!