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[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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