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Issue #16203: Add re.fullmatch() function and regex.fullmatch() method,

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which anchor the pattern at both ends of the string to match.

Original patch by Matthew Barnett.
This commit is contained in:
Serhiy Storchaka 2013-11-23 23:20:30 +02:00
parent 3ed82c55a8
commit 32eddc1bbc
8 changed files with 168 additions and 37 deletions
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30
Lib/test/test_re.py
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@ -349,6 +349,36 @@ class ReTests(unittest.TestCase):
(None, 'b', None))
self.assertEqual(pat.match('ac').group(1, 'b2', 3), ('a', None, 'c'))
def test_re_fullmatch(self):
# Issue 16203: Proposal: add re.fullmatch() method.
self.assertEqual(re.fullmatch(r"a", "a").span(), (0, 1))
for string in "ab", S("ab"):
self.assertEqual(re.fullmatch(r"a|ab", string).span(), (0, 2))
for string in b"ab", B(b"ab"), bytearray(b"ab"), memoryview(b"ab"):
self.assertEqual(re.fullmatch(br"a|ab", string).span(), (0, 2))
for a, b in "\xe0\xdf", "\u0430\u0431", "\U0001d49c\U0001d49e":
r = r"%s|%s" % (a, a + b)
self.assertEqual(re.fullmatch(r, a + b).span(), (0, 2))
self.assertEqual(re.fullmatch(r".*?$", "abc").span(), (0, 3))
self.assertEqual(re.fullmatch(r".*?", "abc").span(), (0, 3))
self.assertEqual(re.fullmatch(r"a.*?b", "ab").span(), (0, 2))
self.assertEqual(re.fullmatch(r"a.*?b", "abb").span(), (0, 3))
self.assertEqual(re.fullmatch(r"a.*?b", "axxb").span(), (0, 4))
self.assertIsNone(re.fullmatch(r"a+", "ab"))
self.assertIsNone(re.fullmatch(r"abc$", "abc\n"))
self.assertIsNone(re.fullmatch(r"abc\Z", "abc\n"))
self.assertIsNone(re.fullmatch(r"(?m)abc$", "abc\n"))
self.assertEqual(re.fullmatch(r"ab(?=c)cd", "abcd").span(), (0, 4))
self.assertEqual(re.fullmatch(r"ab(?<=b)cd", "abcd").span(), (0, 4))
self.assertEqual(re.fullmatch(r"(?=a|ab)ab", "ab").span(), (0, 2))
self.assertEqual(
re.compile(r"bc").fullmatch("abcd", pos=1, endpos=3).span(), (1, 3))
self.assertEqual(
re.compile(r".*?$").fullmatch("abcd", pos=1, endpos=3).span(), (1, 3))
self.assertEqual(
re.compile(r".*?").fullmatch("abcd", pos=1, endpos=3).span(), (1, 3))
def test_re_groupref_exists(self):
self.assertEqual(re.match('^(\()?([^()]+)(?(1)\))$', '(a)').groups(),
('(', 'a'))