问题描述
我有一个QApplication，根据命令行参数，有时实际上没有GUI窗口，但只是运行没有GUI。在这种情况下，如果CTRL-C命中，我想优雅地关闭它。基本上我的代码看起来像这样：
I have a QApplication that, depending on command line parameters, sometimes doesn't actually has a GUI window, but just runs without GUI. In this case, I want to shut it down gracefully if CTRL-C was hit. Basically my code looks like this:
int main(int argc, char* argv[])
{
QApplication app(argc, argv);
... // parse command line options
if (no_gui) {
QObject::connect(&app, SIGNAL(unixSignal(int)),
&app, SLOT(quit()));
app.watchUnixSignal(SIGINT, true);
app.watchUnixSignal(SIGTERM, true);
}
...
return app.exec();
}
但是，这不起作用。 CTRL-C似乎被捕获（应用程序不会被杀死），但它也不退出。
However, this does not work. CTRL-C seems to be caught (the application doesn't get killed), but it also doesn't exit. What am I missing?
推荐答案
由于没有记录， QApplication :: watchUnixSignal 不应该使用。并且，从阅读代码，它使用glib事件分派器（这是Linux上的默认值）时无法正常工作。
As it isn't documented, QApplication::watchUnixSignal shouldn't be used. And, from reading the code, it will not work properly when using the glib event dispatcher (which is the default on Linux).
但是，一般来说，可以在Qt应用程序中安全地捕获Unix信号，您只需要自己写一点代码。在文档中甚至有一个例子 - 。
However, in general you can safely catch Unix signals in Qt applications, you just have to write a bit of the code yourself. There is even an example in the documentation - Calling Qt Functions From Unix Signal Handlers.
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