`QKeySequence::match()`何时返回`QKeySeque::PartialMatch`

Question

哪些序列部分匹配？

我认为1与Ctrl+1部分匹配，但事实并非如此。

Answer 1

取自源代码

QKeySequence::SequenceMatch QKeySequence::matches(const QKeySequence &seq) const {
uint userN = count(),
seqN = seq.count();
if (userN > seqN)
return NoMatch;
// If equal in length, we have a potential ExactMatch sequence,
// else we already know it can only be partial.
SequenceMatch match = (userN == seqN ? ExactMatch : PartialMatch);
for (uint i = 0; i < userN; ++i) {
QKeyCombination userKey = (*this)[i],
sequenceKey = seq[i];
if (userKey != sequenceKey)
return NoMatch;
}
return match;
}

看来

1. 被比较的序列必须长于源
2. 比较序列必须从源开始

所以如果你有，

#include "mainwindow.h"
#include <QApplication>
#include <QDebug>
#include <QKeySequence>
int main(int argc, char *argv[])
{
QApplication a(argc, argv);
QKeySequence ks_1{Qt::CTRL, Qt::Key_X, Qt::Key_Y};
QKeySequence ks_2{Qt::CTRL};
QKeySequence ks_3{Qt::CTRL, Qt::Key_X};
QKeySequence ks_4{Qt::CTRL, Qt::Key_X, Qt::Key_Y};
QKeySequence ks_5{Qt::CTRL, Qt::Key_X, Qt::Key_Y, Qt::Key_Z};
qDebug() << ks_1.count() << ks_2.count() << ks_1.matches(ks_2);
qDebug() << ks_1.count() << ks_3.count() << ks_1.matches(ks_3);
qDebug() << ks_1.count() << ks_4.count() << ks_1.matches(ks_4);
qDebug() << ks_1.count() << ks_5.count() << ks_1.matches(ks_5);
return a.exec();
}

ks_4将是完全匹配，而ks_5将是部分匹配。注意，长度是使用count()获取的，其来源为：

int QKeySequence::count() const
{
return int(std::distance(d->key, std::find(d->key, d->key + QKeySequencePrivate::MaxKeyCount, 0)));
}

因此，如果前面的例子被声明为：

QKeySequence ks_1{Qt::CTRL + Qt::Key_X + Qt::Key_Y};
QKeySequence ks_2{Qt::CTRL};
QKeySequence ks_3{Qt::CTRL + Qt::Key_X};
QKeySequence ks_4{Qt::CTRL + Qt::Key_X + Qt::Key_Y};
QKeySequence ks_5{Qt::CTRL + Qt::Key_X + Qt::Key_Y + Qt::Key_Z};

所有情况都不匹配，因为count()对以上所有情况都返回1。