嵌套 DirectConnection 信号导致段错误

问题描述

这是我的此处未回答的问题.与下面的代码完全一样，它会因段错误而崩溃(将其复制/粘贴到您的系统中并运行它).当我从一个或两个信号构造函数调用中删除 type=QtCore.Qt.DirectConnection 时(因此使用 QtCore.Qt.AutoConnection 代替)，一切都按照它应该的方式运行:出现一个小部件，显示五个进度条在无限循环中填满然后清空.

This is a follow-up to my unanswered question here. The code, exactly as it is below crashes with a segfault (copy/paste it into your system and run it). When I remove type=QtCore.Qt.DirectConnection from one or both of the signal constructor calls (thereby using QtCore.Qt.AutoConnection instead), everything runs the way it should: A widget appears, showing five progress bars filling up then emptying in an endless loop.

from PySide import QtCore, QtGui
import time

class Worker(QtCore.QThread):
    sig_worker_update_progress = QtCore.Signal(int, int)

    def __init__(self, thread_id, *args, **kwargs):
        super(Worker, self).__init__(*args, **kwargs)
        self.thread_id = thread_id
        self.stop_requested = False

    def slot_interrupt(self):
        self.stop_requested = True

    def run(self):
        progress = 0
        while(True):
            self.sig_worker_update_progress.emit(self.thread_id, progress % 100)
            progress += 1
            if self.stop_requested:
                break
            else:
                time.sleep(0.1)


class Controller(QtCore.QObject):
    sig_controller_update_progress = QtCore.Signal(int, int)

    def __init__(self, num_workers, *args, **kwargs):
        super(Controller, self).__init__(*args, **kwargs)

        self.workers = []
        for i in range(num_workers):
            self.workers.append(Worker(i))
            self.workers[i].sig_worker_update_progress.connect(
                self.slot_worker_update_progress,
                type=QtCore.Qt.DirectConnection)
        for worker in self.workers:
            worker.start()

    def slot_worker_update_progress(self, thread_id, progress):
        # Do
        # Stuff
        self.sig_controller_update_progress.emit(thread_id, progress)


class Monitor(QtGui.QWidget):
    def __init__(self, num_workers, *args, **kwargs):
        super(Monitor, self).__init__(*args, **kwargs)
        main_layout = QtGui.QVBoxLayout()
        self.setLayout(main_layout)
        self.progress_bars = []

        for _ in range(num_workers):
            progress_bar = QtGui.QProgressBar()
            main_layout.addWidget(progress_bar)
            self.progress_bars.append(progress_bar)

        self.controller  = Controller(num_workers)
        self.controller.sig_controller_update_progress.connect(
            self.slot_controller_update_progress,
            type=QtCore.Qt.DirectConnection)

    def slot_controller_update_progress(self, thread_id, progress):
        self.progress_bars[thread_id].setValue(progress)


if __name__ == "__main__":
    app = QtGui.QApplication([])
    monitor = Monitor(5)
    monitor.show()
    app.exec_()

为什么使用两个嵌套的 DirectConnection 信号会导致段错误?如果 Qt 不希望您这样做，为什么没有给出更多信息错误?

Why does using two nested DirectConnection signals cause a segfault? If Qt does not want you to do that, why isn't a more informative error given?

我使用的是 PySide v1.2.2，它封装了 Qt 4.8 框架.

I am using PySide v1.2.2 which wraps the Qt 4.8 framework.

推荐答案

我找到了一个令人满意的解释 这里.显然，发出 DirectConnection 类型的信号相当于直接函数调用.因此，当两个信号都是 DirectConnect 时，GUI 毕竟是在 Worker 线程上更新的.正如我在其他问题中所提到的，线程不允许改变GUI.问题不在于嵌套 DirectConnection 本身.

I found a satisfying explanation here. Apparently, emitting a signal of type DirectConnection is equivalent to a direct function call. So the GUI is after all updated on a Worker thread when both signals are DirectConnect-ed. As mentioned on my other question, threads are not allowed to change the GUI. The issue is NOT with nesting DirectConnections per se.

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