What is causing “unbound method __init__() must be called with instance as first argument” from this Python code?

问题

I have this class:

from threading import Thread 
import time

class Timer(Thread): 
    def __init__(self, interval, function, *args, **kwargs): 
        Thread.__init__() 
        self.interval = interval 
        self.function = function 
        self.args = args 
        self.kwargs = kwargs 
        self.start()

    def run(self): 
        time.sleep(self.interval) 
        return self.function(*self.args, **self.kwargs) 

and am calling it with this script:

    import timer 
    def hello():
        print \"hello, world
    t = timer.Timer(1.0, hello)
    t.run()

and get this error and I can't figure out why: unbound method __init__() must be called with instance as first argument

回答1:

You are doing:

Thread.__init__() 

Use:

Thread.__init__(self) 

Or, rather, use super()

回答2:

This is a frequently asked question at SO, but the answer, in brief, is that the way you call your superclass's constructor is like:

super(Timer,self).__init__()

回答3:

First, the reason you must use:

Thread.__init__(self)

instead of

Thread.__init__()

is because you are using the class name, and not an object (an instance of the class), so you cannot call a method in the same way as an object.

Second, if you are using Python 3, the recommended style for invoking a super class method from a sub class is:

super().method_name(parameters)

Although in Python 3 is possible to use:

SuperClassName.method_name(self, parameters)

It is an old style of syntax that is not the prefer style.

回答4:

You just need to pass 'self' as an argument to 'Thread.init'. After that, it works on my machines.

来源：https://stackoverflow.com/questions/1615148/what-is-causing-unbound-method-init-must-be-called-with-instance-as-first