AuroraMiddleware/qt5base-lts
1
0
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity

Bugfix in QDateTimeParser's findTextEntry()

Browse Source 
If a later month-or-day were to have a name that's a prefix of an
earlier one's name, the code would have selected the longer name as
best match when the text matched is the shorter name, simply because
it found that one first.  (Found, on Turkish Cuma(rtesi)? in Thiago's
recent new test, by reversing the loop that iterated the list.)

Make an exact match win and a match of a full name beat any prefix
match of the same length.

Change-Id: I8d954b83ccc25e4f47af2e558036d714685cef5e
Reviewed-by: Thiago Macieira <thiago.macieira@intel.com>
This commit is contained in:
Edward Welbourne 2017-02-28 12:03:46 +01:00
parent 326f1fdb7d
commit 9ed389bf15
1 changed files with 10 additions and 4 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

14
src/corelib/tools/qdatetimeparser.cpp
View File

@ -1248,9 +1248,12 @@ end:
\internal
\brief Returns the index in \a entries with the best prefix match to \a text
Scans \a entries looking for an entry overlapping \a text as much as possible.
Records the length of overlap in *used (if \a used is non-NULL) and the first
entry that overlapped this much in *usedText (if \a usedText is non-NULL).
Scans \a entries looking for an entry overlapping \a text as much as possible
(an exact match beats any prefix match; a match of the full entry as prefix of
text beats any entry but one matching a longer prefix; otherwise, the match of
longest prefix wins, earlier entries beating later on a draw). Records the
length of overlap in *used (if \a used is non-NULL) and the first entry that
overlapped this much in *usedText (if \a usedText is non-NULL).
*/
static int findTextEntry(const QString &text, const QVector<QString> &entries, QString *usedText, int *used)
{
@ -1267,9 +1270,12 @@ static int findTextEntry(const QString &text, const QVector<QString> &entries, Q
int i = 0;
while (i < limit && text.at(i) == name.at(i).toLower())
++i;
if (i > bestCount) {
// Full match beats an equal prefix match:
if (i > bestCount || (i == bestCount && i == name.size())) {
bestCount = i;
bestMatch = n;
if (i == name.size() && i == text.size())
break; // Exact match, name == text, wins.
}
}
if (usedText && bestMatch != -1)