Qt get application state without event listener

Question

I need to know if a key (let's say r) is pressed when the main() start. See :

int main(int argc, char *argv[])
{
    if(R is pressed)
    {} // Do a few things

    // Do amazing stuff whatever happened
    return a.exec();
}

But I can't find a way to do it for all platforms (win, mac, lin), the only thing I found is a trick for windows : Using GetKeyState() which is not very satisfying...

It's very unusual goal. Are you sure you want to implement such strange behavior and make your users use it? Maybe you should use more common ways, e.g. command line options or GUI. If you still want to do this, please describe why you need it. — Pavel Strakhov, Sep 01 '14 at 22:43
@PavelStrakhov I need to implement a reset option on startup which can be used by a noob — Thomas Ayoub, Sep 02 '14 at 07:52

score 2 · Accepted Answer · answered Sep 02 '14 at 13:22

If you want to check modifier keys (shift, control, alt) you can use QGuiApplication::queryKeyboardModifiers() :

int main(int argc, char *argv[])
{
    if(QGuiApplication::queryKeyboardModifiers().testFlag(Qt::ShiftModifier))
    {} // Do a few things

    // Do amazing stuff whatever happened
    return a.exec();
}

score 1 · Answer 2 · answered Sep 02 '14 at 05:08

You can use QxtGlobalShortcut which is a class in Qxt. It provides a global shortcut aka "hotkey" and triggers even if the application is not active :

#include <QApplication>

#include <QxtGlobalShortcut>
#include <QTimer>

int main(int argc, char *argv[])
{
    QApplication a(argc, argv);

    QEventLoop loop;

    QxtGlobalShortcut* shortcut = new QxtGlobalShortcut();
    shortcut->setShortcut(QKeySequence("R"));
    QObject::connect(shortcut, SIGNAL(activated()), shortcut, SLOT(setDisabled()));
    QObject::connect(shortcut, SIGNAL(activated()), &loop,SLOT(quit()));
    QTimer::singleShot(300,&loop,SLOT(quit()));

    loop.exec();

    if(!shortcut->isEnabled())
    {
        //R is pressed
    }

    ...


    return a.exec();
}

Here we wait for maximum 300 milliseconds to check if the key is pressed. When the activated() signal is emitted, the shortcut is disabled and the event loop gets quit. Otherwise the timeout for the timer is activated and the event loop quits.

After getting the source for Qxt and compiling, you should add these to your .pro file :

CONFIG  += qxt

QXT     += core gui

This sounds like a good solution but can't we achieve the same with only `Qt`? — Thomas Ayoub, Sep 02 '14 at 07:50

Qt get application state without event listener

2 Answers2