在Python中实现回调 - 将可调用的引用传递给当前函数 [英] Implementing a callback in Python - passing a callable reference to the current function

问题描述

我想在Python中为几个工作者实现 Observable 模式，并遇到了以下有用的代码段：

I want to implement the Observable pattern in Python for a couple of workers, and came across this helpful snippet:

class Event(object):
    pass

class Observable(object):
    def __init__(self):
        self.callbacks = []
    def subscribe(self, callback):
        self.callbacks.append(callback)
    def fire(self, **attrs):
        e = Event()
        e.source = self
        for k, v in attrs.iteritems():
            setattr(e, k, v)
        for fn in self.callbacks:
            fn(e)

源：这里

据我所知，为了 subscribe ，我需要传递一个回调给在 fire 上调用的函数。如果调用函数是一个 class 方法，大概我可以使用 self ，但在没有这个 - 我可以直接得到一个回调，可以用于 self.callbacks.append（callback） bit？

As i understand it, in order to subscribe, I would need to pass a callback to the function that is going to be called on fire. If the calling function was a class method, presumably I could have used self, but in the absence of this - how could I directly get a callback that can be useful for the self.callbacks.append(callback) bit?

推荐答案

任何定义的函数都可以通过简单的使用它的名字来传递，而不用在你要用来调用它的那一端添加（） ：

Any defined function can be passed by simply using its name, without adding the () on the end that you would use to invoke it:

def my_callback_func(event):
    # do stuff

o = Observable()
o.subscribe(my_callback_func)

---

其他示例用法：

---

Other example usages:

class CallbackHandler(object):
    @staticmethod
    def static_handler(event):
        # do stuff

    def instance_handler(self, event):
        # do stuff

o = Observable()

# static methods are referenced as <class>.<method>
o.subscribe(CallbackHandler.static_handler)

c = CallbackHandler()
# instance methods are <class instance>.<method>
o.subscribe(c.instance_handler)

# You can even pass lambda functions
o.subscribe(lambda event: <<something involving event>>)

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