Question

我正在撰写一份功能校正员,将申请转换为职能的第一个理由。如果我只一次贬低我的职能,但如果我两次贬低他们,我就错了。下面是一些显示这一问题的法典,它是一米法典的简化版本。我排除了转换法,以免脱离问题。

from inspect import getargspec
from functools import wraps

def dec(id):
    def _dec(fn):
        @wraps(fn)
        def __dec(*args, **kwargs):
            if len(args):
                return fn(args[0], *args[1:], **kwargs)
            else:
                first_arg = getargspec(fn).args[0]
                new_kwargs = kwargs.copy()
                del new_kwargs[first_arg]
                return fn(kwargs[first_arg], **new_kwargs)
        return __dec
    return _dec

@dec(1)
def functionWithOneDecorator(a, b, c):
    print "functionWithOneDecorator(a = %s, b = %s, c = %s)" % (a, b, c)

@dec(1)
@dec(2)
def functionWithTwoDecorators(a, b, c):
    print "functionWithTwoDecorators(a = %s, b = %s, c = %s)" % (a, b, c)

functionWithOneDecorator(1, 2, 3)
functionWithOneDecorator(1, b=2, c=3)
functionWithOneDecorator(a=1, b=2, c=3)
functionWithOneDecorator(c=3, b=2, a=1)

functionWithTwoDecorators(1, 2, 3)
functionWithTwoDecorators(1, b=2, c=3)
functionWithTwoDecorators(a=1, b=2, c=3)
functionWithTwoDecorators(c=3, b=2, a=1)

在我管理上述法典时,我取得了以下成果:

functionWithOneDecorator(a = 1, b = 2, c = 3)
functionWithOneDecorator(a = 1, b = 2, c = 3)
functionWithOneDecorator(a = 1, b = 2, c = 3)
functionWithOneDecorator(a = 1, b = 2, c = 3)
functionWithTwoDecorators(a = 1, b = 2, c = 3)
functionWithTwoDecorators(a = 1, b = 2, c = 3)
IndexError: list index out of range

这是因为,当第二名矫正员检查其职能时,它会发现争论姓名,而且由于它正在对一位矫正员进行校正,而只使用 *和 **。

我可以想办法解决这一问题,这些问题将在上述法典中发挥作用,但如果某项职能与我的勋章者和第三者脱节,则仍然会中断。是否有办法解决这一问题? 还是有更好的办法实现同样的结果?

<>Update: 参看

from decorator import decorator

def dec(id):
    @decorator
    def _dec(fn, *args, **kwargs):
        return fn(args[0], *args[1:], **kwargs)
    return _dec

@dec(1)
def functionWithOneDecorator(a, b, c):
    print "functionWithOneDecorator(a = %s, b = %s, c = %s)" % (a, b, c)

@dec(1)
@dec(2)
def functionWithTwoDecorators(a, b, c):
    print "functionWithTwoDecorators(a = %s, b = %s, c = %s)" % (a, b, c)

functionWithOneDecorator(1, 2, 3)
functionWithOneDecorator(1, b=2, c=3)
functionWithOneDecorator(a=1, b=2, c=3)
functionWithOneDecorator(c=3, b=2, a=1)

functionWithTwoDecorators(1, 2, 3)
functionWithTwoDecorators(1, b=2, c=3)
functionWithTwoDecorators(a=1, b=2, c=3)
functionWithTwoDecorators(c=3, b=2, a=1)

更清洁和工作!

Answer 1

问题是,贵重职务的签名不是原件的签名。在 decorator 模块的帮助中,确实作了很好的解释。你可以解决你的问题。基本上,你应当使用签字保留标记,以便第二代腐蚀器看到与第一个标识相同的签名。

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