如何使用模板与QT信号和插槽？ [英] How to use templates with QT signals and slots?

问题描述

我想在我的程序中使用信号和插槽，但不幸的是，它们应该用于传输几种不同的数据类型（例如QString，double等），我不想写20个不同的插槽，只是因为我需要每个数据类型一个。但是当我要声明一个像

I want to use signals and slots in my program, but unfortunately they should be used for transmitting several different data types (e.g. QString, double, etc.) and I don't want to write twenty different slots just because I need one for each data type. But when I want to declare a slot like

template <typename t>
void Slot1(t data);

QT告诉我，不可能在信号和插槽中使用模板。有解决方法吗？

QT tells me that it is not possible to use templates in signals and slots. Is there a workaround? Or can my approach simply improved?

推荐答案

您可以使用新的信号和插槽语法：

You can do this with new signals and slots syntax:

QSlider *slid = new QSlider;
QLineEdit *lne = new QLineEdit;

connect(slid,&QSlider::valueChanged,this,&MainWindow::random);
connect(lne,&QLineEdit::textChanged,this,&MainWindow::random);
lne->show();
slid->show();

插槽：

void MainWindow::random(QVariant var)
{
    qDebug() << var;
}

输出：

QVariant(int, 11) 
QVariant(int, 12) 
QVariant(int, 13) 
QVariant(int, 14) 
QVariant(int, 16) 
QVariant(QString, "c") 
QVariant(QString, "cv") 
QVariant(QString, "cvb") 
QVariant(QString, "cvbc") 
QVariant(QString, "cvbcv")

http://qt-project.org/wiki/New_Signal_Slot_Syntax

如果存在隐式
转换（例如从QString到QVariant），可以自动转换类型

Possibility to automatically cast the types if there is implicit conversion (e.g. from QString to QVariant)

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