“self”在另一个函数（self）中声明函数（self）时丢失了一些内容

Question

请注意以下代码（Win10上的python3.6，PyCharm），函数thread0(self)作为线程成功启动，但是thread1(self)似乎与thread0(self)的设置方式不同。self.thread0可以，但self.thread1不行。self.thread1中的self在它的类函数中没有thread1，但是它有__init__()中的所有内容。事实上，在PyCharm中，参数self甚至没有在def thread1(self):行中突出显示。我的理解是，像foo(self)这样的语法会将foo（）添加为self所指类的成员。你知道吗

既然我们在这里，我就无法解释为什么启动thread0的try-catch块中的代码也失败了，可能与线程的特定语法要求有关？你知道吗

我有一种感觉，像这样嵌套使用self可能是不推荐的。但在我的实际代码中，我确实需要线程声明位于新进程中而不是main()，这样这些线程就可以共享该进程的同一个python记录器。你知道吗

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

您忘了理解self是jussst变量名，表示什么是另一回事，要使代码正常工作，您只需为变量选择另一个名称，请看：

Important edit

您忘记了在名为t4的线程中以方法为目标

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)

现在您得到了6个输出，例如：

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