为什么我可以将一个命名参数列表 - 但不是未命名的参数 - 传递给这个装饰器?

时间:2017-06-06 20:21:03

标签: python decorator python-decorators

此问题与How pass unknown list of unnamed arguments to a python decorator?不重复。我在这里问一个不同但相关的问题。

我创建了一个python decorator my_decorator方法,如下所示。我希望这个装饰器接受一个未知的参数列表:

#!/usr/bin/env python
from functools import wraps

class A:
    def my_decorator(self, func=None, *args, **kwargs):
        print "Hello World2!"
        print 'args = {}'.format(args)
        print 'kwargs = {}'.format(kwargs)
        def inner_function(decorated_function):
            def wrapped_func(*fargs, **fkwargs):
                print "Hello World3!"
                return decorated_function(*fargs, **fkwargs)
            return wrapped_func

        if func:
            return inner_function(func)
        else:
            return inner_function

class B:
    my_a = A()

    @my_a.my_decorator(a1="Yolo", b1="Bolo")
    def my_func(self):
         print "Hello World1!"

my_B = B()
my_B.my_func()

此代码完美无缺:

Hello World2!
args = ()
kwargs = {'a1': 'Yolo', 'b1': 'Bolo'}
Hello World3!
Hello World1!

但是,现在,我没有将命名参数传递给@my_a.my_decorator,而是想传递这样的未命名参数:@my_a.my_decorator('Yolo', 'Bolo')并且它失败了:

Hello World2!
args = ('Bolo',)
kwargs = {}
Hello World3!
Traceback (most recent call last):
  File "./decorator_test.py", line 20, in <module>
    class B:
  File "./decorator_test.py", line 23, in B
    @my_a.my_decorator('Yolo', 'Bolo')
  File "./decorator_test.py", line 12, in wrapped_func
    return decorated_function(*fargs, **fkwargs)
TypeError: 'str' object is not callable

我该如何解决这个问题?

1 个答案:

答案 0 :(得分:1)

def my_decorator(self, *args, **kwargs):
    [skip]
    if 'func' in kwargs:
        return inner_function(kwargs.pop('func'))
    else:
        return inner_function