Question

我有这种主要(Qt)申请。

int main(int argc, char *argv[])
{
    QApplication app(argc, argv);
    MainWindow   mainWin;

    ... A separate, non-GUI thread is launched here

    mainWin.Init();
    mainWin.show();

    app.exec();
}

这是在主线之前创造的另一种深层。 Win需要知道何时可以开始与主Win沟通。但自从主妇使用以来信号、位置、时间等,在事件发生之前,它没有真正准备好岩石(通过外部()。

我的问题是:在事件爆发时,是否有信号或事件?

考虑这一点。在主Win.Init(Init)中,你可以创造像Qtimer这样的东西,甚至叫.......开始()把它拉开。但是,它实际上获得了胜利,在被召唤之前就触发事件。因此,我需要知道事件发生的时间。

Answer 1

You can send a signal to your window before the exec() call. This will place an entry in app s signal queue. When exec() is running, the signal will be delivered and your window will know that the event loop is running.

使用<代码>Qtimer: Shot(0, &mainWin, SLOT(onEventLoopStarted()));,与你的窗口类别习惯位置连接。

Answer 2

Since emitted signals don t get lost when the event loop is not yet running, your thread may not necessarily need to know when your window is ready.
Your thread could start sending signals to the window right away but it will only receive signals from the window when the event loop is running.

Answer 3

你可以按以下顺序这样做:

QApplication app(argc, argv);
Mainwinwdow mainWin;
QThread yourThread;

//connect the signals from the thread to the mainWin here

mainWin.Init();
mainWin.show();

yourThread.start();

return app.exec();

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