python - 从装饰器访问拥有装饰方法的类

Question

我正在为必须检查父方法(我正在装饰的类的父类中同名的方法)的方法编写装饰器。

示例(来自 PEP 318 的第四个示例):

def returns(rtype):
    def check_returns(f):
        def new_f(*args, **kwds):
            result = f(*args, **kwds)
            assert isinstance(result, rtype), \
                   "return value %r does not match %s" % (result,rtype)
            return result
        new_f.func_name = f.func_name
        # here I want to reach the class owning the decorated method f,
        # it should give me the class A
        return new_f
    return check_returns

class A(object):
    @returns(int)
    def compute(self, value):
        return value * 3

所以我正在寻找代码来代替 # here I want...

谢谢。

Answer 1

作为bobince said it ，您不能访问周围的类，因为在调用装饰器时，该类还不存在。如果您需要访问类和基础的完整字典，您应该考虑 metaclass :

__metaclass__

This variable can be any callable accepting arguments for name, bases, and dict. Upon class creation, the callable is used instead of the built-in type().

基本上，我们将 returns 装饰器转换为仅告诉元类在类构造上做一些魔术的东西:

class CheckedReturnType(object):
    def __init__(self, meth, rtype):
        self.meth = meth
        self.rtype = rtype

def returns(rtype):
    def _inner(f):
        return CheckedReturnType(f, rtype)
    return _inner

class BaseInspector(type):
    def __new__(mcs, name, bases, dct):
        for obj_name, obj in dct.iteritems():
            if isinstance(obj, CheckedReturnType):
                # do your wrapping & checking here, base classes are in bases
                # reassign to dct
        return type.__new__(mcs, name, bases, dct)

class A(object):
    __metaclass__ = BaseInspector
    @returns(int)
    def compute(self, value):
        return value * 3

注意我还没有测试过这段代码，如果我应该更新这个请留下评论。

有一些articles on metaclasses由非常值得推荐的 David Mertz 撰写，在这种情况下您可能会发现它很有趣。