Python动态装饰器-为什么要这么多包装？ [英] Python dynamic decorators - why so many wraps?

问题描述

因此，我对Python装饰器还是一种新手-以前曾经使用过它们，但我从来没有自己做过。我正在阅读本教程（该特定段落），但我没有似乎不明白为什么我们需要三个功能级别？我们为什么不能做这样的事情？

So I'm still kind of new to Python decorators - I've used them before, but I've never made my own. I'm reading this tutorial (that particular paragraph) and I don't seem to understand why do we need three levels of functions? Why can't we do something like this:

def decorator(func, *args, **kwargs):
    return func(*args,**kwargs)

谢谢：）

推荐答案

那么，如果您在函数上调用该装饰器会发生什么？

Well, what would happen if you called that decorator on a function?

@decorator
def foo(): pass

此代码将立即调用我们不想要的foo。装饰器被调用，它们的返回值替换该函数。就像说

This code would immediately call foo, which we don't want. Decorators are called and their return value replaces the function. It's the same as saying

def foo(): pass
foo = decorator(foo)

因此，如果我们有一个调用foo的装饰器，我们可能希望有一个函数返回一个调用foo的函数-

So if we have a decorator that calls foo, we probably want to have a function that returns a function that calls foo -- that function that it returns will replace foo.

def decorator(f):
    def g(*args, **kwargs):
        return f(*args, **kwargs)
    return g

现在，如果我们想将选项传递给装饰器，则无法像您的示例中那样通过函数并排准确地将它们并排传递。没有语法。因此，我们定义了一个返回参数化装饰器的函数。

Now, if we want to pass options to the decorator, we can't exactly pass them ins ide-by-side with the function like in your example. There's no syntax for it. So we define a function that returns a parameterized decorator. The decorator it returns will be a closure.

def argument_decorator(will_I_call_f):
    def decorator(f):
        def g(*args, **kwargs):
            if will_I_call_f: return f(*args, **kwargs)
        return g
    return decorator

所以我们可以做到

decorator = argument_decorator(True)
@decorator
def foo(): pass

Python提供了方便的语法，可以在其中内联函数调用：

And Python offers the convenience syntax where you inline the function call:

@argument_decorator(True)
def foo(): pass

所有这些都是非装饰符

def foo(): pass
foo = argument_decorator(True)(foo)

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