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
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`.)
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)
--- /dev/null
+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)``.
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) ?
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) ?
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) ?
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) ?
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) ?
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) ?