Question

我知道QSignalMapper适合这样的条件：

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()));

现在我想要实现3个滑块，所有滑块都有一个SLOT按钮：

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SIGNAL和SLOT之间的一致规则相矛盾。这里有解决方法吗？我正在使用Qt4。

Answer 1

QSignalMapper不是要将信号从信号发送到插槽，而是让信号接收器知道“谁”是那个或者使用什么数据。如果你需要知道值和发送者，你可以使用一些内部类映射，或使用QObject * mapper然后将QObject *转换为滑块。

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);

这是插槽体：

void YourClass::SetSlice(QWidget *wgt)
{
    QSlider * slider = qobject_cast<QSlider *>(wgt);

    if(slider) {
        SetSlice(slider->value());
    }
}

带有信号参数和额外参数的QSignalMapper

1 个答案: