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Re: Bug? ( () == [] ) != ( ().__eq__([]) )

Started byChris Angelico <rosuav@gmail.com>
First post2013-08-05 00:06 +0100
Last post2013-08-05 00:06 +0100
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  Re: Bug? ( () == [] ) != ( ().__eq__([]) ) Chris Angelico <rosuav@gmail.com> - 2013-08-05 00:06 +0100

#51943 — Re: Bug? ( () == [] ) != ( ().__eq__([]) )

FromChris Angelico <rosuav@gmail.com>
Date2013-08-05 00:06 +0100
SubjectRe: Bug? ( () == [] ) != ( ().__eq__([]) )
Message-ID<mailman.195.1375658082.1251.python-list@python.org>
On Sun, Aug 4, 2013 at 11:35 PM, Markus Rother <python@markusrother.de> wrote:
> Hello,
>
> The following behaviour seen in 3.2 seems very strange to me:
>
> As expected:
>>>> () == []
> False
>
> However:
>>>> ().__eq__([])
> NotImplemented
>>>> [].__eq__(())
> NotImplemented

You don't normally want to be calling dunder methods directly. The
reasoning behind this behaviour goes back to a few things, including a
way to handle "1 == Foo()" where Foo is a custom type that implements
__eq__; obviously the integer 1 won't know whether it's equal to a Foo
instance or not, so it has to defer to the second operand to get a
result. This deferral is done by returning NotImplemented, which is an
object, and so is true by default. I don't see any particular reason
for it to be false, as you shouldn't normally be using it; it's more
like a "null" state, it means "I don't know if we're equal or not". If
neither side knows whether they're equal, then they're presumed to be
unequal, but you can't determine that from a single call to __eq__.

ChrisA

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