New submission from ferno:
Document in question is:
http://docs.python.org/3.3/reference/datamodel.html#object.__iadd__
The documentation states,
to execute the statement x += y, where x is an instance of a class that has
an __iadd__() method, x.__iadd__(y) is called.
However, this doesn't appear to be strictly true. According to this, the
following well-known example:
>>> a = (1, [2, 3])
>>> a[1] += [4, 5]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: 'tuple' object does not support item assignment
>>> a
(1, [2, 3, 4, 5])
should give the same behaviour as:
>>> a = (1, [2, 3])
>>> a[1].__iadd__([4, 5])
[2, 3, 4, 5]
>>> a
(1, [2, 3, 4, 5])
However, this snippet DOES give the identical behaviour:
>>> a = (1, [2, 3])
>>> a[1] = a[1].__iadd__([4, 5])
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: 'tuple' object does not support item assignment
>>> a
(1, [2, 3, 4, 5])
which leads me to suggest that this line of the documentation should be
adjusted to read:
to execute the statement x += y, where x is an instance of a class that has
an __iadd__() method, x = x.__iadd__(y) is called.
This fix would incidentally harmonise with the documentation for
operator.iadd() et al.,
(http://docs.python.org/3.3/library/operator.html#operator.iadd), which had a
similar problem but was corrected following issue 7259
(http://bugs.python.org/issue7259), now reading:
a = iadd(a, b) is equivalent to a += b.
----------
assignee: docs@python
components: Documentation
messages: 205920
nosy: docs@python, ferno
priority: normal
severity: normal
status: open
title: __iadd__() doc not strictly correct
versions: Python 2.6, Python 2.7, Python 3.1, Python 3.2, Python 3.3, Python
3.4, Python 3.5
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Python tracker <[email protected]>
<http://bugs.python.org/issue19953>
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