python - 当在另一个函数(self)中声明函数(self)时，“self”丢失了一些东西

Question

请观察以下代码(Win10上的python 3.6，PyCharm)，函数thread0(self)作为线程成功启动，但是 thread1(self)似乎与thread0(self)不同已设置。 self.thread0很好，但是self.thread1不是。 self在self.thread1没有thread1在其类函数中，但它具有 __init__() 中的所有内容。事实上，在 PyCharm 中，参数 self def thread1(self): 行中甚至没有突出显示。我的理解是像foo(self)这样的语法会将 foo() 添加为 self 所指向的类的成员.

既然说到这里，我无法解释为什么启动thread0的try-catch block 中的代码也失败了，也许这与线程的具体语法要求有关？

我有一种感觉，嵌套使用self这样可能不推荐。但在我的实际代码中，我确实需要在新进程中声明线程，而不是 main() ，以便这些线程可以共享该进程的同一个 python 记录器。

import threading
import multiprocessing
from time import sleep


class exe0(multiprocessing.Process):
    def __init__(self):
        super().__init__()
        self.aaa = 111

        # working syntax for thread0
        t = threading.Thread(
            target=self.thread0,
            daemon=1,
        )
        t.start()

        try:
            # NOT working syntax
            t = threading.Thread(
                target=thread0,
                args=(self,),
                daemon=1,
            )
            t.start()
            sleep(1)
        except Exception as e:
            print(e)

    def thread0(self):
        print(type(self))

    def run(self):

        # working syntax for thread1
        def thread1(self):
            print(type(self))
            print(self.aaa)

        t = threading.Thread(
            target=thread1,
            args=(self,),
            daemon=1,
        )
        t.start()
        sleep(1)

        try:
            # NOT working syntax
            t = threading.Thread(
                target=self.thread1,
                daemon=1,
            )
            t.start()
            sleep(1)
        except Exception as e:
            print(e)


if __name__ == '__main__':
    multiprocessing.freeze_support()
    e = exe0()
    e.daemon = 1
    e.start()
    sleep(2)

# output:
'''
<class '__main__.exe0'>
name 'thread0' is not defined
<class '__mp_main__.exe0'>
111
'exe0' object has no attribute 'thread1'
'''

Answer 1

你忘了明白 self 只是一个变量名，代表的是另一回事，为了让你的代码正常工作，你只需为你的变量选择另一个名字，看看:

Important edit

您忘记将名为t4的线程中的方法作为目标

import threading
import multiprocessing
from time import sleep


class exe0(multiprocessing.Process):
    def __init__(self):
        super().__init__()
        self.aaa = 111

        t1 = threading.Thread(
            target=self.thread0,
            daemon=1,
        )
        t1.start()

        try:
            t2 = threading.Thread(
                target=self.thread0, #here I removed the other parameter
                daemon=1,
            )
            t2.start()
            sleep(1)
        except Exception as e:
            print(e)

    def thread0(self):
        print(type(self))


    def run(self):
        def thread1(s): #here you can see the name doesn't matter
            print(type(s)) #here you can see the name doesn't matter
            print(s.aaa)


        t3 = threading.Thread(
            target=thread1(self), 
            daemon=1,
        )
        t3.start()
        sleep(1)

        try:
            t4 = threading.Thread(
                target=thread1(self), #here there is no need of the parameter
                daemon=1,
            )
            t4.start()
            sleep(1)
        except Exception as e:
            print(e)


multiprocessing.freeze_support()
e = exe0()
e.daemon = 1
e.start()
sleep(2)

现在您获得了 6 个输出，例如:

<class 'exe0'>
<class 'exe0'>
<class 'exe0'>
111
<class 'exe0'>
111