在我的工作线程为“主”线程的Qt应用中，如何与GUI线程通信

Question

Normally in a Qt app, one would create a QApplication and then call it's exec method to start the GUI thread.

I've got a slightly different arrangement... I have my own thread (ie, not a QThread) and it is blocking on I/O. At times, it will need to send information to the GUI thread.

Is there a safe and easy way to do this?

I looked here: http://qt-project.org/doc/qt-5/threads-technologies.html but nothing really stuck me as easy or straightforward.

I am using QT5.3 on OS X

Answer 1

Beware, if your thread is not QThread or it does not run Qt event loop you will not be able to use signal-slot queued connections since the queued events will not be processed actually. In such cases you can opt for classical semaphore/mutex synchronization as you would do in non-Qt application.

Answer 2

You can set up a slot on a QObject and use QMetaObject::invokeMethod using the default or queued connection to ensure the method is called on the thread of the QObject.

You can also call a signal of the QObject directly:

class MessagePasser: public QObject{
Q_OBJECT

public:
MessagePasser(QObject* p):QObject(p){}

signals:
void message(QByteArray);

}

and in your thread you can simply call message on a MessagePasser object when you need to send some info:

MessagePasser* passer;


QByteArray data;
//get data with blocking operation
passer->message(data);//call signal and it will pass it to the gui thread correctly through the connected slots

Answer 3

The only way to communicate with the GUI from a thread safely is via signals and slots, and you cannot do that from a custom thread. A QThread has the abilty to connect its signals to a GUIs main thread.

Answer 4

The easiest way to do this is to create a QObject derived class on your main thread whose responsibility is to communicate events from the main thread using signals. Just make sure you instantiate it on the main thread. Your main thread does not have to be a QThread .

Qt will automatically use a queued connection when the source and the receiver are on different threads, which means the slot will be executed on the receiver's thread.