超类__init__没有认出它的kwargs

Question

我正在尝试使用an answer to another question提供的StoppableThread类：

import threading

# Technique for creating a thread that can be stopped safely
# Posted by Bluebird75 on StackOverflow
class StoppableThread(threading.Thread):
    """Thread class with a stop() method. The thread itself has to check
    regularly for the stopped() condition."""

    def __init__(self):
        super(StoppableThread, self).__init__()
        self._stop = threading.Event()

    def stop(self):
        self._stop.set()

    def stopped(self):
        return self._stop.isSet()

但是，如果我运行类似：

st = StoppableThread(target=func)

我明白了：

  

TypeError：__init__()有一个意外的关键字参数'target'

可能是对如何使用它的疏忽。

Answer 1

StoppableThread类不会在构造函数中将threading.Thread的任何其他参数传递或传递给它。你需要做这样的事情：

class StoppableThread(threading.Thread):
    """Thread class with a stop() method. The thread itself has to check
    regularly for the stopped() condition."""

    def __init__(self,*args,**kwargs):
        super(threading.Thread,self).__init__(*args,**kwargs)
        self._stop = threading.Event()

这会将位置和关键字参数传递给基类。

Answer 2

您要覆盖 init ， init 不会接受任何参数。你应该添加一个“target”参数并将它传递给你的基类构造函数，超级甚至更好地通过* args和* kwargs允许任意参数。

即

def __init__(self,*args,**kwargs):
    super(threading.Thread,self).__init__(*args,**kwargs)
    self._stop = threading.Event()