我们如何使如下代码模拟__getattribute__继承自的方法object？我真的很想修改的行为，__getattribute__但我想从获得自然行为开始。
class Descriptor:
def __get__(*args):
pass
def __set__(*args)
pass
class K:
__getattribute__ = Descriptor()
问问题
48 次
1 回答
-1
是的，我们可以做__getattribute__一个描述符！首先，如果存在一个名为 的整数成员变量x，那么__get__为其编写的描述符方法x应该返回一个整数。同样，因为__getattribute__is 是一个函数，所以__get__for__getattribute__将返回一个函数。最初，让我们__get__返回一个名为 的愚蠢函数foo，这样我们就可以运行程序看看发生了什么：
class Descriptor:
def __get__(*args):
print("ENTERING __get__")
print("Type(arg) for args passed into __get__: ", end="")
print(", ".join(map(lambda x: type(x).__name__, args)))
def foo(*args):
print("ENTERING `foo`")
print("args passed into foo: ", end="")
print(", ".join(repr(arg) for arg in args))
print("LEAVING `foo`")
return "I AM THE RETURN VALUE OF FOO"
return foo
class K:
__getattribute__ = Descriptor()
现在，我们尝试以下方法：
instance = K()
x = instance.x
以下是在控制台 ( stdout) 上看到的语句：
ENTERING __get__
Type(arg) for args passed into __get__: Descriptor, K, type
ENTERING `foo`
args passed into foo: 'x'
LEAVING `foo`
请注意，foo不通过通常的self参数接收实例对象。相反，只foo接收属性名称'x'
通常，以下两段代码产生相同的最终结果：
x = instance.x
x = K.__getattribute__(instance, 'x')
让我们尝试运行K.__getattribute__(instance, 'x')：
ENTERING __get__
Type(arg) for args passed into __get__: Descriptor, NoneType, type
ENTERING `foo`
args passed into foo: <__main__.K object at 0x01AFE0B8>, 'x'
LEAVING `foo`
input to__get__和 input tofoo实际上和以前不同了。
+---------------------+-----------------------------+----------------------------+
| instance.x | __get__( | foo(<<string `x`>>) |
| | <<descriptor instance>>, | |
| | <<instance of K class>>, | |
| | <<K class itself>> | |
| | ) | |
+---------------------+-----------------------------+----------------------------+
| K.__getattribute__( | __get__( | foo( |
| instance, 'x' | <<descriptor instance>>, | <<instance of K class>>,|
| ) | <<None>>, | <<string `x`>> |
| | <<K class itself>>, | |
| | | |
+---------------------+-----------------------------+----------------------------+
因此，我们希望__get__如下所示：
def __get__(descriptor, Kinstance, Kclass):
if Kinstance:
"""
return a function, `f`, which accepts as its only input
the string name of an attribute.
leftChild = f('leftChild')
`f` is supposed to return that attribute
"""
else: # Kinstance == None
"""
return a function, `f`, which accepts as input two items:
1) instance of some class
2)the string name of an attribute.
inst = Klass()
leftChild = f(inst, 'leftChild')
`f` is supposed to return that attribute
"""
如果您想要 的默认行为__getattribute__，以下几乎可以工作：
class Descriptor:
def __get__(descriptor, self, cls):
if self:
lamby = lambda attrname:\
object.__getattribute__(self, attrname)
return lamby
else: # self == None
return object.__getattribute__
class K:
__getattribute__ = Descriptor()
全局变量问题
在self命名的 lambda 函数中作为全局变量lamby是非常危险的。当 lambda 函数最终被调用时，self将与self定义 lambda 函数时存在的不同。考虑以下示例：
color = "white"
get_fleece_color = lambda shoop:\
shoop + ", whose fleece was as " + color + " as snow."
color是 lambda 函数内部的成员变量get_fleece_color。get_fleece_color定义的时间color是"white"，但这可能会改变：
print(get_fleece_color("Igor"))
# [... many lines of code later...]
color = "pink polka-dotted"
print(get_fleece_color("Igor's cousin, 3 times removed"))
输出是：
Igor, whose fleece was white as snow.
Igor's cousin, 3 times removed Igor, whose fleece was as pink polka-dotted as snow.
def使用关键字而不是make 的函数lambda同样危险。
lamby = lambda attrname:\
object.__getattribute__(self, attrname)
def lamby(attrname):
object.__getattribute__(self, attrname)
我们希望使用在定义self的同时存在的值lamby，而不是使用被调用self时存在的值lamby。有几种方法可以做到这一点，其中 3 种如下所示：
避免全局变量解决方案 1
lamby = lambda attrname, *, self=self:\
object.__getattribute__(self, attrname)
避免全局变量解决方案 2
class Descriptor:
def __get__(descriptor, self, cls):
lamby = object.__getattribute__
if self: # self != None
Method = type(descriptor.__get__)
lamby = Method(object.__getattribute__, self)
return lamby
避免全局变量解决方案 3
class SamMethod:
def __init__(self, func, arg):
self.func = func
self.arg = arg
def __call__(self, *args, **kwargs):
return self.func(self.arg, *args, **kwargs)
class Descriptor:
def __get__(descriptor, self, cls):
lamby = object.__getattribute__
if self: # self != None
lamby = SamMethod(object.__getattribute__, self)
return lamby
class K:
__getattribute__ = Descriptor()
def foo(self):
pass
关于全局变量问题的最后评论
解决方案2存在一个问题，即method类构造函数的签名可能会在未来版本的 python 中发生变化。解决方案3是我最喜欢的，但请随意选择您自己的。
遗产
继承工作得很好：
class SamMethod:
def __init__(self, func, arg):
self.func = func
self.arg = arg
def __call__(self, *args, **kwargs):
return self.func(self.arg, *args, **kwargs)
#############################################################################
class Descriptor:
def __get__(descriptor, self, cls):
print(
"MESSAGE"
)
lamby = object.__getattribute__
if self: # self != None
lamby = SamMethod(object.__getattribute__, self)
return lamby
###############################################################################
class ParentOfKlaus:
__getattribute__ = Descriptor()
class Klaus(ParentOfKlaus):
def foo(self):
print("Klaus.foo")
class ChildOfKlaus(Klaus):
def __init__(self):
self.x = 99
def foo(self):
print("ChildOfKlaus.foo")
instance = ChildOfKlaus()
y = instance.x
print(y) # 99
instance.foo()
控制台输出为：
MESSAGE
99
MESSAGE
ChildOfKlaus.foo
为什么要创建__getattribute__ 描述符而不是覆盖__getattribute__？
一个优点是我们可以访问传递给描述符__get__方法的类参数。注意__get__被称为 as__get__(descriptor, Kinstance, Kclass)我们通常不能直接访问Kclass，但是如果__getattribute__写成描述符，那么它可以访问Kclass
于 2019-11-12T01:43:38.090 回答