[Python-Dev] == on object tests identity in 3.x
Xavier Morel
python-dev at masklinn.net
Mon Jul 7 17:58:39 CEST 2014
On 2014-07-07, at 13:22 , Andreas Maier <andreas.r.maier at gmx.de> wrote:
> While discussing Python issue #12067 (http://bugs.python.org/issue12067#msg222442), I learned that Python 3.4 implements '==' and '!=' on the object type such that if no special equality test operations are implemented in derived classes, there is a default implementation that tests for identity (as opposed to equality of the values).
>
> The relevant code is in function do_richcompare() in Objects/object.c.
>
> IMHO, that default implementation contradicts the definition that '==' and '!=' test for equality of the values of an object.
>
> Python 2.x does not seem to have such a default implementation; == and != raise an exception if attempted on objects that don't implement equality in derived classes.
That's incorrect on two levels:
1. What Terry notes in the bug comments is that because all Python 3
types inherit from object this can be done as a default __eq__/__ne__,
in Python 2 the fallback is encoded in the comparison framework
(PyObject_Compare and friends):
http://hg.python.org/cpython/file/01ec8bb7187f/Objects/object.c#l756
2. Unless comparison methods are overloaded and throw an error it will
always return either True or False (for comparison operator), never throw.
> I'd like to gather comments on this issue, specifically:
>
> -> Can someone please elaborate what the reason for that is?
>
> -> Where is the discrepancy between the documentation of == and its default implementation on object documented?
>
> To me, a sensible default implementation for == on object would be (in Python):
>
> if v is w:
> return True;
> elif type(v) != type(w):
> return False
> else:
> raise ValueError("Equality cannot be determined in default implementation")
Why would comparing two objects of different types return False but
comparing two objects of the same type raise an error?
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