[docs] [issue27646] yield from expression can be any iterable
Terry J. Reedy
report at bugs.python.org
Thu Jul 28 14:30:50 EDT 2016
New submission from Terry J. Reedy:
"When yield from <expr> is used, it treats the supplied expression
as a subiterator. All values produced by that subiterator ...".
To me "treats..expression as a subiterator" means that the expression must *be* an iterator, such as returned by iter or calling a generator function. Hence I was surprised upon reading "yield from <non-iterator iterable>" in stdlib code.
I confirmed that this usage is correct by trying
>>> def g():
yield from (1,2)
>>> i = g()
>>> next(i), next(i)
and then reading the PEP380 Formal Semantics, which begins with "_i = iter(EXPR)". Hence I suggest the following replacement for the quote above:
"When yield from <expr> is used, the expression must be an iterable.
A subiterator is obtained with iter(<expr>). All values produced
by that subiterator ...".
Note that 'subiterator' is spelled in the following sentences 'underlying iterable' (which I am not sure I like) and 'sub-iterator' (and 'sub-generator'). I think we should be consistent for at least the two short 'yield from' paragraphs.
assignee: docs at python
nosy: docs at python, terry.reedy
stage: patch review
title: yield from expression can be any iterable
versions: Python 3.5, Python 3.6
Python tracker <report at bugs.python.org>
More information about the docs