题
所以,这个作品:
·H
public slots:
void addMenu(QString passedName);
signals:
void clicked(const QString &text);
的.cpp
signalMapper = new QSignalMapper(this);
signalMapper->setMapping(button, QString("passed_value"));
connect(button, SIGNAL(clicked()), signalMapper, SLOT(map()));
connect(signalMapper, SIGNAL(mapped(QString)), this, SLOT(addMenu(QString)));
现在,我想传递一个QMAP
·H
public slots:
void addMenu(QMap < QString, QString > map);
signals:
void clicked(const QMap < QString, QString > &map);
的.cpp
//map is defined above
signalMapper = new QSignalMapper(this);
signalMapper->setMapping(button, map);
connect(button, SIGNAL(clicked()), signalMapper, SLOT(map()));
connect(signalMapper, SIGNAL(mapped(QObject*)), this, SLOT(addMenu(QMap < QString, QString >)));
编辑:我还尝试添加一个typedef,但仍然得到相同的错误
·H
public:
typedef QMap < QString, QString > passedMapType;
public slots:
void addMenu(passedMapType map);
signals:
void clicked(passedMapType map);
的.cpp
passedMapType passedMap;
passedMap.insert(QString("key"), QString("value"));
signalMapper = new QSignalMapper(this);
signalMapper->setMapping(button, passedMap);
connect(button, SIGNAL(clicked()), signalMapper, SLOT(map()));
connect(signalMapper, SIGNAL(mapped(QObject*)), this, SLOT(addMenu(passedMapType));
....
addMenu(passedMapType passedMap) {
}
解决方案
“现在,我想传递一个QMAP
这不是一个普通的信号/槽或宏扩展问题,而是QSignalMapper的限制。你可以不QMAP传递给setMapping,仅INT,QString的,QWidget的*和* QObject的。请参阅 QSignalMapper文档。
connect(signalMapper, SIGNAL(mapped(QObject*)), this, SLOT(addMenu(passedMapType));
这将不能工作,因为签名是不兼容的:QObject的*与QMAP。
我会做什么:保持一个QMAP
QMap<QWidget*, QMap<QString, QString> > map; //member of the class
//when creating the buttons:
for each button b:
signalMapper->setMapping( b, b );
map.insert( b, someMapForThisButton );
connect( signalMapper, SIGNAL(mapped(QWidget*)), this, SLOT(addMenuForButton(QWidget*)) );
//the slot:
void addMenuForButton(QWidget* w) {
const QMap<QString, QString> m = map.value( w );
create menu...
}
其他提示
使用typedef。我的感觉是两个的QString模板参数之间的逗号呈宏观膨胀的问题。
作为一个很好的解决办法,你可以使用QObject的。 QObject的具有嵌入式QMAP内。下面是如何不言而喻:
void sender() {
...
...
QObject *data = new QObject(0);
data->setProperty("Name","xxxx");
data->setProperty("Address","yyyy");
//You can also send QImage
QImage image;
...
data->setProperty("Image",image);
emit dataReady(data);
}
signals:
void dataReady(QObject*);
public slots:
void receiver(QObject *data) {
QString Name = data->property("Name").toString();
QString Address = data->property("Address").toString();
QImage image = data->property("Image").value<QImage>();
data->deleteLater();
}
这是一个纯粹的猜测,但你可能需要使用指针类型呢?我从未有过很幸运通过引用传递与&
public slots:
void addMenu(QMap<QString, QString> *map);
signals:
void clicked(QMap<QString, QString> *map);
在信号和槽必须具有相同的参数类型(它的一个函数调用,各种各样的),所以这条线是错误的:
connect(signalMapper, SIGNAL(mapped(QObject*)), this, SLOT(addMenu(passedMapType));
您可能需要重新考虑你的逻辑有点使这项工作。