在 python 上使用字典作为装饰器

Question

有一些方法可以使用 dict 作为装饰器 python 使用默认的 " @" 语法?

喜欢这样的东西：

from functools import wraps
def deco(f):
    @wraps(f)
    def wrapper(*args, **kwargs):
        print("WRAPPER")
        return f(*args, **kwargs)
    return wrapper

X={ "deco": deco }

# Working
@deco
def f():
    pass

# Not Working: SyntaxError: invalid syntax
@X["deco"]
def y():
    pass

f()
y()

如果更改为在 dict 上使用 get()，这个工作，但是不要像这样：

# Not working: SyntaxError: invalid syntax
@X.get("deco").require(http_exception=403)
def y():
    pass

这个语法有一些方法可以工作or/and为什么不工作？

Answer 1

语法是有意限制的。见 PEP-0318:

The decorator statement is limited in what it can accept -- arbitrary expressions will not work. Guido preferred this because of a gut feeling.

The current syntax also allows decorator declarations to call a function that returns a decorator:

email the PEP links 说：

I have a gut feeling about this one. I'm not sure where it comes from, but I have it. It may be that I want the compiler to be able to recognize certain decorators.

So while it would be quite easy to change the syntax to @test in the future, I'd like to stick with the more restricted form unless a real use case is presented where allowing @test would increase readability. (@foo().bar() doesn't count because I don't expect you'll ever need that).

因此，该语言的 creator/lead 贡献者决定，允许任何表达式用作装饰器都会导致代码难以阅读。

也从来没有必要。你总是可以做

my_deco = X.get("deco").require(http_exception=403)
@my_deco
def somefunc():
    ...

在 python 上使用字典作为装饰器

Using a dict as decorators on python

python

function

decorator

wrapper