fdo#72219: Fix for corruption of symbols in docx

Issue:
OUString uses UTF-16, so for a Unicode surrogate character there are 2
values stored, not just 1.
So we are getting assert failure in "rtl_uString_iterateCodePoints" method.

erAck: Underlying cause was that the dictionary breakiterator misused UTF-16 positions as Unicode code point positions.

Change-Id: I923485f56c2d879b63687adaea2b489a3479991c
Reviewed-on: https://gerrit.libreoffice.org/6955
Reviewed-by: Eike Rathke <erack@redhat.com>
Tested-by: Eike Rathke <erack@redhat.com>
This commit is contained in:
Rohit Deshmukh 2013-12-06 15:42:53 +05:30 committed by Eike Rathke
parent 45b72633d1
commit d8fd158759
2 changed files with 22 additions and 3 deletions

View file

@ -45,7 +45,7 @@ public:
#endif
void testKhmer();
void testJapanese();
void testChinese();
CPPUNIT_TEST_SUITE(TestBreakIterator);
CPPUNIT_TEST(testLineBreaking);
CPPUNIT_TEST(testGraphemeIteration);
@ -64,6 +64,7 @@ public:
CPPUNIT_TEST(testLao);
#endif
CPPUNIT_TEST(testJapanese);
CPPUNIT_TEST(testChinese);
CPPUNIT_TEST_SUITE_END();
private:
uno::Reference<i18n::XBreakIterator> m_xBreak;
@ -938,6 +939,22 @@ void TestBreakIterator::testJapanese()
}
}
void TestBreakIterator::testChinese()
{
lang::Locale aLocale;
aLocale.Language = "zh";
aLocale.Country = "CN";
i18n::Boundary aBounds;
{
const sal_Unicode CHINESE[] = { 0x6A35, 0x6A30, 0x69FE, 0x8919, 0xD867, 0xDEDB };
OUString aTest(CHINESE, SAL_N_ELEMENTS(CHINESE));
aBounds = m_xBreak->getWordBoundary(aTest, 4, aLocale,
i18n::WordType::DICTIONARY_WORD, true);
CPPUNIT_ASSERT(aBounds.startPos == 4 && aBounds.endPos == 6);
}
}
void TestBreakIterator::setUp()
{
BootstrapFixtureBase::setUp();

View file

@ -387,9 +387,11 @@ Boundary xdictionary::getWordBoundary(const OUString& rText, sal_Int32 anyPos, s
if (u_isWhitespace(ch))
i--;
}
boundary.endPos = boundary.startPos;
rText.iterateCodePoints(&boundary.endPos, aCache.wordboundary[i]);
rText.iterateCodePoints(&boundary.startPos, aCache.wordboundary[i-1]);
boundary.endPos += aCache.wordboundary[i];
boundary.startPos += aCache.wordboundary[i-1];
} else {
boundary.startPos = anyPos;
if (anyPos < len) rText.iterateCodePoints(&anyPos, 1);