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gh-124130: Fix a bug in matching regular expression \B in empty string (GH-127007)

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Serhiy Storchaka 2025-01-02 14:11:21 +02:00 • committed by GitHub
parent 8d16919a06
commit a3711d1541
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5 changed files with 15 additions and 25 deletions
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7
Doc/library/re.rst
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@ -572,11 +572,8 @@ character ``'$'``.
Word boundaries are determined by the current locale
if the :py:const:`~re.LOCALE` flag is used.
.. note::
Note that ``\B`` does not match an empty string, which differs from
RE implementations in other programming languages such as Perl.
This behavior is kept for compatibility reasons.
.. versionchanged:: next
``\B`` now matches empty input string.
.. index:: single: \d; in regular expressions

4
Doc/whatsnew/3.14.rst
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@ -245,6 +245,10 @@ Other language changes
making it a :term:`generic type`.
(Contributed by Brian Schubert in :gh:`126012`.)
* ``\B`` in :mod:`regular expression <re>` now matches empty input string.
Now it is always the opposite of ``\b``.
(Contributed by Serhiy Storchaka in :gh:`124130`.)
* iOS and macOS apps can now be configured to redirect ``stdout`` and
``stderr`` content to the system log. (Contributed by Russell Keith-Magee in
:gh:`127592`.)

13
Lib/test/test_re.py
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@ -978,18 +978,15 @@ class ReTests(unittest.TestCase):
self.assertIsNone(re.fullmatch(br".+\B", b"abc", re.LOCALE))
self.assertIsNone(re.fullmatch(r".+\B", "ьюя"))
self.assertTrue(re.fullmatch(r".+\B", "ьюя", re.ASCII))
# However, an empty string contains no word boundaries, and also no
# non-boundaries.
# However, an empty string contains no word boundaries.
self.assertIsNone(re.search(r"\b", ""))
self.assertIsNone(re.search(r"\b", "", re.ASCII))
self.assertIsNone(re.search(br"\b", b""))
self.assertIsNone(re.search(br"\b", b"", re.LOCALE))
# This one is questionable and different from the perlre behaviour,
# but describes current behavior.
self.assertIsNone(re.search(r"\B", ""))
self.assertIsNone(re.search(r"\B", "", re.ASCII))
self.assertIsNone(re.search(br"\B", b""))
self.assertIsNone(re.search(br"\B", b"", re.LOCALE))
self.assertTrue(re.search(r"\B", ""))
self.assertTrue(re.search(r"\B", "", re.ASCII))
self.assertTrue(re.search(br"\B", b""))
self.assertTrue(re.search(br"\B", b"", re.LOCALE))
# A single word-character string has two boundaries, but no
# non-boundary gaps.
self.assertEqual(len(re.findall(r"\b", "a")), 2)

4
Misc/NEWS.d/next/Library/2024-11-19-10-46-57.gh-issue-124130.OZ_vR5.rst Normal file
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@ -0,0 +1,4 @@
Fix a bug in matching regular expression ``\B`` in empty input string.
Now it is always the opposite of ``\b``.
To get an old behavior, use ``(?!\A\Z)\B``.
To get a new behavior in old Python versions, use ``(?!\b)``.

12
Modules/_sre/sre_lib.h
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@ -42,8 +42,6 @@ SRE(at)(SRE_STATE* state, const SRE_CHAR* ptr, SRE_CODE at)
return ((void*) ptr == state->end);
case SRE_AT_BOUNDARY:
if (state->beginning == state->end)
return 0;
thatp = ((void*) ptr > state->beginning) ?
SRE_IS_WORD((int) ptr[-1]) : 0;
thisp = ((void*) ptr < state->end) ?
@ -51,8 +49,6 @@ SRE(at)(SRE_STATE* state, const SRE_CHAR* ptr, SRE_CODE at)
return thisp != thatp;
case SRE_AT_NON_BOUNDARY:
if (state->beginning == state->end)
return 0;
thatp = ((void*) ptr > state->beginning) ?
SRE_IS_WORD((int) ptr[-1]) : 0;
thisp = ((void*) ptr < state->end) ?
@ -60,8 +56,6 @@ SRE(at)(SRE_STATE* state, const SRE_CHAR* ptr, SRE_CODE at)
return thisp == thatp;
case SRE_AT_LOC_BOUNDARY:
if (state->beginning == state->end)
return 0;
thatp = ((void*) ptr > state->beginning) ?
SRE_LOC_IS_WORD((int) ptr[-1]) : 0;
thisp = ((void*) ptr < state->end) ?
@ -69,8 +63,6 @@ SRE(at)(SRE_STATE* state, const SRE_CHAR* ptr, SRE_CODE at)
return thisp != thatp;
case SRE_AT_LOC_NON_BOUNDARY:
if (state->beginning == state->end)
return 0;
thatp = ((void*) ptr > state->beginning) ?
SRE_LOC_IS_WORD((int) ptr[-1]) : 0;
thisp = ((void*) ptr < state->end) ?
@ -78,8 +70,6 @@ SRE(at)(SRE_STATE* state, const SRE_CHAR* ptr, SRE_CODE at)
return thisp == thatp;
case SRE_AT_UNI_BOUNDARY:
if (state->beginning == state->end)
return 0;
thatp = ((void*) ptr > state->beginning) ?
SRE_UNI_IS_WORD((int) ptr[-1]) : 0;
thisp = ((void*) ptr < state->end) ?
@ -87,8 +77,6 @@ SRE(at)(SRE_STATE* state, const SRE_CHAR* ptr, SRE_CODE at)
return thisp != thatp;
case SRE_AT_UNI_NON_BOUNDARY:
if (state->beginning == state->end)
return 0;
thatp = ((void*) ptr > state->beginning) ?
SRE_UNI_IS_WORD((int) ptr[-1]) : 0;
thisp = ((void*) ptr < state->end) ?