免责声明这只是元编程的练习,它没有实际意义 目的
我在函数对象上分配了__getitem__
和__getattr__
方法,但是
没有效果......
def foo():
print "foo!"
foo.__getitem__ = lambda name: name
foo.__getattr__ = lambda name: name
foo.baz = 'baz'
我们可以为某个功能分配属性的完整性检查:
>>> foo.baz
'baz'
纯。 “神奇的吸气者”怎么样?
>>> foo.bar
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: 'function' object has no attribute 'bar'
>>> foo['foo']
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: 'function' object is not subscriptable
>>> getattr(foo, 'bar')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: 'function' object has no attribute 'bar'
是否可以在函数对象上使用“魔术吸气剂”?
答案 0 :(得分:6)
都能跟得上!将__getitem__
分配给实例不适用于任何类型的对象:
>>> class A(object):
... pass
...
>>> a = A()
>>> a.__getattr__ = lambda name: name
>>> a.foo
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: 'A' object has no attribute 'foo'
您无法在内置函数类型中定义__getattr__
:
>>> import types
>>> types.FunctionType.__getitem__ = lambda name: name
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: can't set attributes of built-in/extension type 'function'
您无法继承types.FunctionType
:
>>> import types
>>> class F(types.FunctionType):
... pass
...
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: Error when calling the metaclass bases
type 'function' is not an acceptable base type
答案 1 :(得分:3)
至少在新式类(它是Python 3中唯一的类,以及你应该在Python 2中使用的类)中,Python只在类(及其祖先)上寻找魔术方法,而不是在实例上。 Docs here
当然,您无法修改函数类型,也无法从中派生。但是,正如您所发现的,任何具有__call__()
方法的类都会生成可调用的实例,因此这是实现它的方法。
答案 2 :(得分:1)
AHHA!使用__call__
,并将功能包装在F()
class F(object):
def __init__(self, fn):
self.__dict__['fn'] = fn
def __call__(self, *args, **kwargs):
return self.fn(*args, **kwargs)
def __getitem__(self, name):
return name
def __getattr__(self, name):
return name
>>> foo = F(foo)
>>> f.bar
'bar'
>>> f['foo']
'foo'
>>> foo()
foo!