Question

考虑以下情况：

class Meta(type):
    def shadowed(cls):
        print "Meta.shadowed()"
    def unshadowed(cls):
        print "Meta.unshadowed()"

class Foo(object):
    __metaclass__ = Meta

    def shadowed(self):
        print "Foo.shadowed()"

我可以调用unshadowed上的绑定方法Foo，它运行正常：

>>> Foo.unshadowed
<bound method Meta.unshadowed of <class '__main__.Foo'>>
>>> Foo.unshadowed()
Meta.unshadowed()

但是，我似乎无法在shadowed上获取绑定方法Foo - 它指向我，而不是必须使用Foo实例调用的未绑定方法：

>>> Foo.shadowed
<unbound method Foo.shadowed>
>>> Foo.shadowed()

Traceback (most recent call last):
  File "<pyshell#45>", line 1, in <module>
    Foo.shadowed()
TypeError: unbound method shadowed() must be called with Foo instance as first argument (got nothing instead)

有没有办法获得<bound method Meta.shadowed of <class '__main__.Foo'>>？

Answer 1

似乎在this answer中找到了关于如何绑定未绑定方法的一个潜在答案（可能不是最好的答案）。所以我们可以这样做：

>>> Meta.shadowed.__get__(Foo, Meta)()
Meta.shadowed()

更好的示范：

class Meta(type):
    def shadowed(cls):
        print "Meta.shadowed() on %s" % (cls.__name__,)
    def unshadowed(cls):
        print "Meta.unshadowed() on %s" % (cls.__name__,)

class Foo(object):
    __metaclass__ = Meta

    def shadowed(self):
        print "Foo.shadowed()"

class Bar(object):
    __metaclass__ = Meta

Bar.unshadowed()                   #Meta.unshadowed() on Bar 
Bar.shadowed()                     #Meta.shadowed() on Bar
Foo.unshadowed()                   #Meta.unshadowed() on Foo
#Foo.shadowed()                    #TypeError    
Meta.shadowed.__get__(Foo, Meta)() #Meta.shadowed() on Foo

当类具有相同名称的方法时，如何获取从元类派生的绑定方法

1 个答案: