New submission from Joseph Fox-Rabinovitz:
The docs for [`operator.index`][1] and `operator.__index__` state that
> Return *a* converted to an integer. Equivalent to `a.__index__()`.
The first sentence is correct, but the second is not. First of all, we have the
data model [docs][2]:
> For custom classes, implicit invocations of special methods are only
> guaranteed to work correctly if defined on an object’s type, not in the
> object’s instance dictionary.
Secondly, we can make a simple counter-example in code:
```
import operator
class A:
def __index__(self): return 0
a = A()
a.__index__ = (lambda self: 1).__get__(a, type(a))
operator.index(a)
```
The result is of course zero and not one.
I believe that the docs should read something more like one of the following to
avoid being misleading:
> Return *a* converted to an integer, if it is already an integral type.
> Return *a* converted to an integer. Equivalent to `type(a).__index__(a)`.
Or a combination of both:
> Return *a* converted to an integer, if it is already an integral type.
> Equivalent to `type(a).__index__(a)`.
[1]: https://docs.python.org/3/library/operator.html#operator.index
[2]: https://docs.python.org/3/reference/datamodel.html#special-method-lookup
----------
assignee: docs@python
components: Documentation
messages: 299195
nosy: docs@python, madphysicist
priority: normal
severity: normal
status: open
title: Inconsistency in documentation of operator.index
type: behavior
versions: Python 2.7, Python 3.3, Python 3.4, Python 3.5, Python 3.6, Python 3.7
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Python tracker <[email protected]>
<http://bugs.python.org/issue31042>
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