QT双击键盘事件

Question

I know I can read single keyboard key pressure from QT C++ program using

void keyPressEvent(QKeyEvent*);

function.

But what can I do, if I want to assign an action on the two consecutive pressures of the same keyboard key, separated by no more than 100ms(or any other fixed timeout)?

Of course, I imply that some another action is assigned to a single pressure of the same key, and I have to execute some another routine if timeout expires without second pressure.

Is there any simpler solution, than creating a second thread with a timer?

I hate an idea of creation one more thread for a such paltry task.

Answer 1

You can use a numPress counter for the number of key pressing. Then start a singleShot right after the first key press for 500 ms. After 500 ms, you can check the number of key pressing and decide which function you should call.

void SO_Qt::keyPressEvent( QKeyEvent* key)
{
    if (key->key() == Qt::Key_K)
    {
        numPress_++;
        if (numPress_ == 1)
        {
            QTimer::singleShot(500, this, SLOT(keyKPressed()));
        }
    }
}

void SO_Qt::keyKPressed()
{
    if (numPress_ == 1) {
        func_1();
    } else if (numPress_ == 2) {
        func_2();
    }
    numPress_ = 0;
}

void SO_Qt::func_1()
{
    QMessageBox::information(this, "1","1");
}

void SO_Qt::func_2()
{
    QMessageBox::information(this, "2","2");
}