QString: fix count(QRegularExpression)

There is an off by one in the implementation of count(): a match
must be attempted even at the very end of the string, because
a 0-length match can happen there. While at it, improve
the documentation on the counter-intuitive behavior of count(),
which doesn't merely count how many times a regexp matches
into a string using ordinary global matching.

[ChangeLog][QtCore][QString] Fixed a corner case when using
QString::count(QRegularExpression), causing an empty in the
last position not to be accounted for in the returned result.

Change-Id: I064497839a96979abfbac2d0a96546ce160bbc46
Pick-to: 5.15 6.0
Reviewed-by: Thiago Macieira <thiago.macieira@intel.com>

1 files changed, 9 insertions, 3 deletions

diff --git a/src/corelib/text/qstring.cpp b/src/corelib/text/qstring.cpp
index 56d9e0a011..02501e00f4 100644
--- a/src/corelib/text/qstring.cpp
+++ b/src/corelib/text/qstring.cpp
@@ -4324,10 +4324,16 @@ bool QString::contains(const QRegularExpression &re, QRegularExpressionMatch *rm
     Returns the number of times the regular expression \a re matches
     in the string.
 
-    This function counts overlapping matches, so in the example
-    below, there are four instances of "ana" or "ama":
+    For historical reasons, this function counts overlapping matches,
+    so in the example below, there are four instances of "ana" or
+    "ama":
 
     \snippet qstring/main.cpp 95
+
+    This behavior is different from simply iterating over the matches
+    in the string using QRegularExpressionMatchIterator.
+
+    \sa QRegularExpression::globalMatch()
 */
 qsizetype QString::count(const QRegularExpression &re) const
 {
@@ -4338,7 +4344,7 @@ qsizetype QString::count(const QRegularExpression &re) const
     qsizetype count = 0;
     qsizetype index = -1;
     qsizetype len = length();
-    while (index < len - 1) {
+    while (index <= len - 1) {
         QRegularExpressionMatch match = re.match(*this, index + 1);
         if (!match.hasMatch())
             break;