如何使用python-decorator包来装饰类方法？ [英] How do I use the python-decorator package to decorate classmethods?

问题描述

我有一个装饰器，我想用它来装饰类方法。在下面的示例中，@mydec装饰器可以正常工作，但是在使用help（）或pydoc时，它不会保留函数签名。为了解决这个问题，我研究了使用@decorator python-decorator软件包：

I'm have a decorator that I want to use to decorate class methods. In the following example, the @mydec decorator works fine on its own, however it does not preserve the function signature when using help() or pydoc. In order to fix this, I looked at using @decorator python-decorator package:

import functools
import decorator


@decorator.decorator
def mydec(func):
    @functools.wraps(func)
    def inner(cls, *args, **kwargs):
        # do some stuff
        return func(cls, *args, **kwargs)
    return inner


class Foo(object):
    @classmethod
    @mydec
    def bar(cls, baz='test', qux=None):
        print (baz, qux)


Foo.bar()

不幸的是，这导致以下异常：

Unfortunately, this results in the following exception:

Traceback (most recent call last):
  File "/tmp/test.py", line 21, in <module>
    Foo.bar()
  File "<string>", line 2, in bar
TypeError: mydec() takes exactly 1 argument (4 given)

推荐答案

您不再需要提供自己的包装器，只需使用 inner 函数上的> @ decorator.decorator ，它需要一个额外的第一个位置参数，该函数包装为：

You do not need to provide your own wrapper anymore, just use @decorator.decorator on the inner function, which takes one extra first positional argument, the function wrapped:

@decorator.decorator
def mydec(func, cls, *args, **kwargs):
    # do some stuff
    return func(cls, *args, **kwargs)

装饰器 package不对装饰使用闭包，而是将包装函数作为参数传递。

The decorator package doesn't use a closure for decorators and instead passes in the wrapped function as an argument.

Demo：

>>> @decorator.decorator
... def mydec(func, cls, *args, **kwargs):
...     # do some stuff
...     return func(cls, *args, **kwargs)
... 
>>> class Foo(object):
...     @classmethod
...     @mydec
...     def bar(cls, baz='test', qux=None):
...         print (baz, qux)
... 
>>> Foo.bar()
('test', None)

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