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

serhiy-storchaka · web-flow · commit a3711d1541c1 · 2025-01-02T12:11:21.000Z
diff --git a/Doc/library/re.rst b/Doc/library/re.rst
@@ -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
 
diff --git a/Doc/whatsnew/3.14.rst b/Doc/whatsnew/3.14.rst
@@ -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`.)
diff --git a/Lib/test/test_re.py b/Lib/test/test_re.py
@@ -978,18 +978,15 @@ def test_word_boundaries(self):
         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)
diff --git a/Misc/NEWS.d/next/Library/2024-11-19-10-46-57.gh-issue-124130.OZ_vR5.rst b/Misc/NEWS.d/next/Library/2024-11-19-10-46-57.gh-issue-124130.OZ_vR5.rst
@@ -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)``.
diff --git a/Modules/_sre/sre_lib.h b/Modules/_sre/sre_lib.h
@@ -42,53 +42,41 @@ 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) ?
             SRE_IS_WORD((int) ptr[0]) : 0;
         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) ?
             SRE_IS_WORD((int) ptr[0]) : 0;
         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) ?
             SRE_LOC_IS_WORD((int) ptr[0]) : 0;
         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) ?
             SRE_LOC_IS_WORD((int) ptr[0]) : 0;
         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) ?
             SRE_UNI_IS_WORD((int) ptr[0]) : 0;
         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) ?
