装饰器执行顺序

问题描述

def make_bold(fn):
    return lambda : "<b>" + fn() + "</b>"

def make_italic(fn):
    return lambda : "<i>" + fn() + "</i>"

@make_bold
@make_italic
def hello():
  return "hello world"

helloHTML = hello()

输出：< b< i>你好世界< / i< / ; / b>

在大多数示例中，我大致了解装饰器及其如何与装饰器一起使用。

I roughly understand about decorators and how it works with one of it in most examples.

在此示例中，有2个。从输出中可以看出，首先执行 @make_italic ，然后执行 @make_bold 。

In this example, there are 2 of it. From the output, it seems that @make_italic executes first, then @make_bold.

这是否意味着对于装饰函数，它将首先运行该函数，然后移至其他装饰器的顶部？先像 @make_italic ，然后是 @make_bold ，而不是相反。

Does this mean that for decorated functions, it will first run the function first then moving towards to the top for other decorators? Like @make_italic first then @make_bold, instead of the opposite.

这意味着它与大多数编程语言中的自顶向下方法的规范不同吗？仅用于这种装饰器吗？还是我错了？

So this means that it is different from the norm of top-down approach in most programming lang? Just for this case of decorator? Or am I wrong?

推荐答案

装饰器包装他们正在装饰的功能。因此， make_bold 装饰了 make_italic 装饰器的结果，该装饰器装饰了 hello 函数。

Decorators wrap the function they are decorating. So make_bold decorated the result of the make_italic decorator, which decorated the hello function.

@decorator 语法实际上只是语法糖。以下内容：

The @decorator syntax is really just syntactic sugar; the following:

@decorator
def decorated_function():
    # ...

实际执行为：

def decorated_function():
    # ...
decorated_function = decorator(decorated_function)

用返回的任何 decorator（）替换原始的 decorated_function 对象。

replacing the original decorated_function object with whatever decorator() returned.

堆叠装饰器重复该过程向外。

Stacking decorators repeats that process outward.

所以您的示例：

@make_bold
@make_italic
def hello():
  return "hello world"

可以扩展为：

def hello():
  return "hello world"
hello = make_bold(make_italic(hello))

现在调用 hello（）时，您正在调用 make_bold（）返回的对象，真的。 make_bold（）返回一个 lambda ，调用函数 make_bold 包装，这是 make_italic（）的返回值，也是返回原始 hello（）的lambda 。扩展所有这些调用，您将得到：

When you call hello() now, you are calling the object returned by make_bold(), really. make_bold() returned a lambda that calls the function make_bold wrapped, which is the return value of make_italic(), which is also a lambda that calls the original hello(). Expanding all these calls you get:

hello() = lambda : "<b>" + fn() + "</b>" #  where fn() ->
    lambda : "<i>" + fn() + "</i>" # where fn() -> 
        return "hello world"

因此输出变为：

"<b>" + ("<i>" + ("hello world") + "</i>") + "</b>"

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