In the manual is says:
in general,
__lt__()and__eq__()are sufficient, if you want the conventional meanings of the comparison operators
But I see the error:
>       assert 2 < three
E       TypeError: unorderable types: int() < IntVar()
when I run this test:
from unittest import TestCase
class IntVar(object):
    def __init__(self, value=None):
        if value is not None: value = int(value)
        self.value = value
    def __int__(self):
        return self.value
    def __lt__(self, other):
        return self.value < other
    def __eq__(self, other):
        return self.value == other
    def __hash__(self):
        return hash(self.value)
class DynamicTest(TestCase):
    def test_lt(self):
        three = IntVar(3)
        assert three < 4
        assert 2 < three
        assert 3 == three
I am surprised that when IntVar() is on the right, __int__() is not being called.  What am I doing wrong?
Adding __gt__() fixes this, but means I don't understand what the minimal requirements are for ordering...
Thanks, Andrew
Python 3.x will never do any type coercions for operators, so __int__() is not used in this context.  The comparison
a < b
will first try to call type(a).__lt__(a, b), and if this returns NotImplemented, it will call type(b).__gt__(b, a).
The quote from the documentation is about making comparisons work for a single type, and the above explanation shows why this would be enough for a single type.
To make your type interact correctly with int, you should either implement all the comparison operator, or use the total_ordering decorator available in Python 2.7 or 3.2.
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With