[New-bugs-announce] [issue22683] bisect index out of bounds issue

Paul Ianas report at bugs.python.org
Tue Oct 21 09:23:12 CEST 2014


New submission from Paul Ianas:

The precondition for all the bisect functions is implemented like this:

    if lo < 0:
        raise ValueError('lo must be non-negative')
    if hi is None:
        hi = len(a)

Now, of course, if hi is given, and hi >= 2 * len(a), then we get an IndexError. In case hi < 0, we always get 0 as a result (even if the element is there).

I think it would be better to treat the hi in the precondition in the same way as the lo parameter: that means, raise a ValueError in case hi has an illegal value.

Disclaimer: of course, it makes no sense to give an illegal argument to that function; still, since lo is treated against illegal values, maybe it's better to do the same for hi.

At the same time, maybe moving the precondition code in a separate function (which raises a ValueError in case precondition is not met) makes more sense, for not repeating the same code in all bisect functions.

A small snippet which reproduces this:

from bisect import bisect_left

a = [1, 2, 3, 4]
idx = bisect_left(a, 2, 0, 10)  # 10 > 2 * 4
print(idx)

----------
components: Library (Lib)
messages: 229750
nosy: Paul.Ianas
priority: normal
severity: normal
status: open
title: bisect index out of bounds issue
type: enhancement
versions: Python 2.7, Python 3.2, Python 3.3, Python 3.4, Python 3.5

_______________________________________
Python tracker <report at bugs.python.org>
<http://bugs.python.org/issue22683>
_______________________________________


More information about the New-bugs-announce mailing list