Question

我有一个控制器类来控制机器人（通过串行接口连接）。该控制器附加到视图。除此之外，我还有一个派生自QThread的线程，它定期读出机器人的状态。

读出状态不得与机器人命令一致，这些命令会从用户界面中触发。因此，我使用QMutexLocker使用互斥锁锁定每个机器人访问，但是如果执行了这样的互斥块，这会导致我的用户界面冻结。

class RobotControl(QObject):
    def __init__(self, view):
        super(RobotControl, self).__init__()
        self.view = view
        self.updatethread = UpdatePositionAndStatus(self.robot)
        self.mutex = QMutex()
        self.connect(self.updatethread, SIGNAL("updateStatus( QString ) "), self.print_error)
        self.updatethread.start()

@pyqtSlot()  
def init_robot(self):
    """
    Initializes the serial interface to the robot interface and checks if
    there is really a robot interface available.
    """
    with QMutexLocker(self.mutex):
        # Open interface
        try:
            index = self.view.robotcontrolui.get_selected_interface_index()
            interface = self.interfaces.item(index).text()
            self.robot = RobotController(interface)
        except DeviceError:
            self.view.error_dlg(self.tr("Couldn't open interface {0}!".format(interface)))
            self.robot = None
            return

        # Check if there is really a robot interface on the selected serial
        # interface with trying to read status byte
        try:
            self.robot.status()
        except DeviceError:
            # In case of failure release interface
            self.close_robot()
            self.view.error_dlg(self.tr("Couldn't initialize robot interface!"))
            return

        self.view.robotcontrolui.bt_open_interface.setEnabled(False)
        self.view.robotcontrolui.bt_close_interface.setEnabled(True)

class UpdatePositionAndStatus(QThread):
    def __init__(self, robot, parent=None):
        QThread.__init__(self, parent) 
        self.robot = robot
        self.mutex = QMutex()
    def run(self):
        """ 
        This function continously reads out the position and the status to for 
        updating it on the userinterface.
        """
        try:
            while True:
                if self.robot is not None:
                    # Do robot communication under a lock
                    self.mutex.lock()
                    (_, rel_pos) = self.robot.read_position()
                    status = self.robot.status()
                    self.mutex.unlock()

                    # Display position and status on userinterface
                    self.view.robotcontrolui.update_position_and_status(rel_pos, status)

                # Wait 1 seccond for next update
                QThread.sleep(1.0)
        except DeviceError:
            # Release lock on robot
            self.mutex.unlock()
            self.emit(SIGNAL("updateStatus( QString )"), self.tr("Error while updating current position and status!"))

触发init方法后，用户界面冻结，程序崩溃：为什么会这样？我怎么能避免这种情况？

Answer 1

很难分辨，因为您的代码示例不完整，但我发现此代码有两个基本问题：

您正在锁定两个不同的QMutex对象。为了使互斥能够正常工作，两个线程必须锁定相同的互斥对象。
您似乎直接在此行的更新线程中与GUI进行交互：
```
self.view.robotcontrolui.update_position_and_status(rel_pos, status)
```
执行GUI操作只能从Qt中的GUI线程完成。这是一个公平的赌注，这是导致你的崩溃。请参阅：http://qt-project.org/doc/qt-4.8/threads.html

QMutexLocker（）导致UI冻结

1 个答案: