Python 装饰器添加了与 Python2 和 Python3 的关键字参数兼容性

Question

我正在编写一个装饰器，它将关键字参数添加到包装的函数中。在本例中，装饰器添加关键字 deco0 和 deco1

def my_decorator(f):
    def inner(arg0, *args, deco0='deco0', deco1='deco1', **kwargs):
        """We assume decoarted functions will take at least one positional arg0"""
        print("Decorated({}, {})".format(deco0, deco1))
        return f(arg0, *args, **kwargs)
    return inner

@my_decorator
def add2(x):
    return x + 2

@my_decorator
def my_mult(x, y):
    return x * y

if __name__ == '__main__':
    print(add2(5))
    print(add2(5, deco0='BLA'))
    print(my_mult(2, 10))
    print(my_mult(2, 10, deco1='BLAM'))

结果符合预期:

Decorated(deco0, deco1)
7
Decorated(BLA, deco1)
7
Decorated(deco0, deco1)
20
Decorated(deco0, BLAM)
20

装饰器返回的inner的函数签名与python2不兼容:

File "./deco.py", line 5
  def inner(arg0, *args, deco0='deco0', deco1='deco1', **kwargs):
                             ^
SyntaxError: invalid syntax

Python2中有没有一种设计模式可以方便装饰器添加关键字参数？或者更好的是，同时兼容 Python2 和 Python3 的模式？

Answer 1

我认为你必须在内部函数内做一些类似修改kwargs的事情:

def inner(arg0, *args, **kwargs):
    deco0 = kwargs.pop('deco0', 'deco0')
    deco1 = kwargs.pop('deco1', 'deco1')
    print("Decorated({}, {})".format(deco0, deco1))
    return f(arg0, *args, **kwargs)