我知道QSignalMapper在这种情况下工作得很好:
QSignalMapper *signalMapper = new QSignalMapper(this);
connect(signalMapper, SIGNAL(mapped(int)), this, SIGNAL(SetSlice(int)));
connect(this->ui->button_1, SIGNAL(slicked()), signalMapper, SLOT(map()));
connect(this->ui->button_2, SIGNAL(clicked()), signalMapper, SLOT(map()));
connect(this->ui->button_3, SIGNAL(clicked()), signalMapper, SLOT(map()));现在我想实现3个滑块都有一个像按钮一样的插槽:
QSignalMapper *signalMapper = new QSignalMapper(this);
connect(signalMapper, SIGNAL(mapped(int)), this, SIGNAL(SetSlice(int)));
connect(this->ui->verticalSlider_1, SIGNAL(valueChanged(int)), signalMapper, SLOT(map()));
connect(this->ui->verticalSlider_2, SIGNAL(valueChanged(int)), signalMapper, SLOT(map()));
connect(this->ui->verticalSlider_3, SIGNAL(valueChanged(int)), signalMapper, SLOT(map()));正如您所看到的,这与信号和插槽之间的一致规则相矛盾。这里有什么变通的方法吗?我正在使用Qt4。
发布于 2013-05-16 15:26:38
QSignalMapper不是将参数从信号发送到插槽,而是让信号接收器知道“谁”是谁或使用了什么数据。如果您需要知道值和发送者,您可以使用一些内部类映射,或者使用QObject *映射器,然后将QObject *转换为slider。
QSignalMapper * mapper = new QSignalMapper(this);
connect(mapper, SIGNAL(map(QWidget *)), this, SLOT(SetSlice(QWidget *)));
mapper->setMapping(this->ui->verticalSlider_1, this->ui->verticalSlider_1);
mapper->setMapping(this->ui->verticalSlider_2, this->ui->verticalSlider_2);
mapper->setMapping(this->ui->verticalSlider_3, this->ui->verticalSlider_3);这里是插槽主体:
void YourClass::SetSlice(QWidget *wgt)
{
QSlider * slider = qobject_cast<QSlider *>(wgt);
if(slider) {
SetSlice(slider->value());
}
}https://stackoverflow.com/questions/16578466
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