线程函数块 main 并不总是

Question

我正在开发一个项目，其中有一个 main class，它创建一个线程来执行一个很长的函数，而 main class（我的 GUI）将继续可用在这个长函数的执行过程中显示一些信息。所以我在 Stack Overflow 上找到了一种方法：

from PyQt5 import QtWidgets, QtCore, QtGui
from PyQt5.QtWidgets import QApplication
import threading
import sys
import time
from PyQt5.QtCore import QObject, pyqtSignal


class Ui_SaftMainWindow(object):
    def setupUi(self, SaftMainWindow):
        SaftMainWindow.setObjectName("SaftMainWindow")
        SaftMainWindow.resize(178, 284)
        icon = QtGui.QIcon()
        icon.addPixmap(QtGui.QPixmap(":/logo/img_Logo-E.png"), QtGui.QIcon.Normal, QtGui.QIcon.Off)
        SaftMainWindow.setWindowIcon(icon)
        self.gridLayout = QtWidgets.QGridLayout(SaftMainWindow)
        self.gridLayout.setObjectName("gridLayout")
        self.frame_all_pb = QtWidgets.QFrame(SaftMainWindow)
        self.frame_all_pb.setMinimumSize(QtCore.QSize(160, 0))
        self.frame_all_pb.setMaximumSize(QtCore.QSize(160, 16777215))
        self.frame_all_pb.setObjectName("frame_all_pb")
        self.gridLayout_7 = QtWidgets.QGridLayout(self.frame_all_pb)
        self.gridLayout_7.setObjectName("gridLayout_7")
        self.pb_stop_simu = QtWidgets.QPushButton(self.frame_all_pb)
        self.pb_stop_simu.setText("STOP")
        self.gridLayout_7.addWidget(self.pb_stop_simu, 5, 0, 1, 1)
        self.pb_start_simu = QtWidgets.QPushButton(self.frame_all_pb)
        self.pb_start_simu.setObjectName("pb_start_simu")
        self.pb_start_simu.setText("START")
        self.gridLayout_7.addWidget(self.pb_start_simu, 4, 0, 1, 1)
        self.gridLayout.addWidget(self.frame_all_pb, 0, 0, 1, 1)

        QtCore.QMetaObject.connectSlotsByName(SaftMainWindow)


class Solver(QObject, object):
    data_changed = pyqtSignal(str, int)

    def __init__(self):
        # Call init function of QObject class for pyQt signal
        QObject.__init__(self)


    def init_parameters(self, input_files, **keys):
        print("Some initialisation")

    def time_step(self):
        # Function with Fortran call
        print("Fortran function")

    def transient_solve(self, arg):
        print("starting")
        t = threading.currentThread()
        i = 1
        while getattr(t, "do_run", True):
            print("working on " + str(i))
            i += 1
            time.sleep(5)
        print("Stopping as you wish.")


class GuiHandler:
    def __init__(self, gui, main_window,):
        # Declarations
        self.gui = gui
        self.main_window = main_window

        self.gui.setupUi(self.main_window)
        self.gui.pb_start_simu.clicked.connect(self.pb_start_simu_clicked)
        self.gui.pb_stop_simu.clicked.connect(self.pb_stop_simu_clicked)

        # Create instance of the second class
        app = Solver()

        # Connect signal to emit signal from second class to ihm later
        app.data_changed.connect(self.myfunction)

        # Initialize some parameters of my second class
        app.init_parameters("", json_file="", output_file="")

        # Create thread object
        self.t = threading.Thread(target=app.transient_solve, args=("task",))

    def pb_start_simu_clicked(self):
        print("Start thread")
        self.t.start()

    def pb_stop_simu_clicked(self):
        print("Stop thread")
        self.t.do_run = False
        self.t.join()

    def myfunction(self, string="this_is_a_test", integer=0):
        print(str(string))
        print(str(integer))


if __name__ == "__main__":
    app = QApplication(sys.argv)
    handler = GuiHandler(Ui_SaftMainWindow(), QtWidgets.QWidget())
    handler.main_window.show()
    sys.exit(app.exec_())

这个简单的例子有效。但是，当我通过调用更大的函数（调用编译为 Python 的 Fortran 文件的 time_step 函数替换睡眠时，GUI 函数被阻止）

知道它是否与我创建和启动线程的方式有关，或者知道我必须 check/find 使我的代码正常工作吗？

我现在不能给你另一个代码示例。

Answer 1

我找到了解决办法，在fortran代码中加入!f2py: threadsafe。

线程函数块 main 并不总是

Threading function block main not all the time

python

multithreading

nonblocking