[issue28019] itertools.count() falls back to fast (integer) mode when step rounds to 1

[Marcos Dione]

New submission from Marcos Dione:

If the `step` parameter for `itertools.count()` rounds to 1 (f.i., 1.1, 1.437643, 1.99999), then it fallsback to fast (integer) mode and increases the counter by 1. Here's an example:

Python 3.6.0a4+ (default:ddc95a9bc2e0+, Sep  8 2016, 14:46:19)
>>> import itertools
>>> for i in itertools.count(1, step=1.5):
...     print(i)
...     if i > 10:
...         break
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----------
components: Library (Lib)
messages: 275007
nosy: StyXman
priority: normal
severity: normal
status: open
title: itertools.count() falls back to fast (integer) mode when step rounds to 1
type: behavior
versions: Python 3.6

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Python tracker <report at bugs.python.org>
<http://bugs.python.org/issue28019>
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