[英]Metaclasses and __slots__?
So, I am reading a bit about metaclasses in Python, and how type()
's three-argument alter-ego is used to dynamically create classes. 所以,我正在阅读Python中的元类,以及如何使用
type()
的三参数alter-ego来动态创建类。 However, the third argument is usually a dict
that initializes the to-be created class' __dict__
variable. 但是,第三个参数通常是一个初始化要创建的类'
__dict__
变量的dict
。
If I want to dynamically create classes based on a metaclass that uses __slots__
instead of __dict__
, how might I do this? 如果我想基于使用
__slots__
而不是__dict__
的元类动态创建类,我该怎么做? Is type()
still used in some fashion along with overriding __new__()
? 是否仍然以某种方式使用
type()
以及重写__new__()
?
As an FYI, I am aware of the proper uses for __slots__
, to save memory when creating large numbers of a class versus abusing it to enforce a form of type-safety. 作为一个FYI,我知道
__slots__
的正确用法,在创建大量类时节省内存而不是滥用它来强制执行类型安全的形式。
Example of a normal (new-style) class that sets __metaclass__
and uses a __dict__
: 设置
__metaclass__
并使用__dict__
的普通(新式)类的__metaclass__
:
class Meta(type):
def __new__(cls, name, bases, dctn):
# Do something unique ...
return type.__new__(cls, name, bases, dctn)
class Foo(object):
__metaclass__ = Meta
def __init__(self):
pass
In the above, type.__new__()
is called and the fourth argument (which becomes the third when actually used) creates a __dict__
in Foo
. 在上面,调用
type.__new__()
,第四个参数(在实际使用时成为第三个参数)在Foo
创建__dict__
。 But if I wanted to modify Meta
to include __slots__
, then I have no dictionary to pass on to type()
's __new__()
function (as far as I know -- I haven't tested any of this yet, just pondering and trying to find some kind of a use-case scenario). 但是如果我想修改
Meta
以包含__slots__
,那么我没有字典可以传递给type()
的__new__()
函数(据我所知 - 我还没有测试过任何这个,只是思考和试图找到某种用例场景)。
Edit: A quick, but untested guess, is to take a dict of the values to be put into the __slots__
variables and pass it to type.__new__()
. 编辑:一个快速但未经测试的猜测是将值放入
__slots__
变量并将其传递给type.__new__()
。 Then add an __init__()
to Meta
that populates the __slots__
variables from the dict. 然后将
__init__()
添加到Meta
,从而填充dict中的__slots__
变量。 Although, I am not certain how that dict would reach __init__()
, because the declaration of __slots__
prevents __dict__
from being created unless __dict__
is defined in __slots__
... 虽然,我不确定该dict将如何到达
__init__()
,因为__slots__
的声明会阻止创建__dict__
,除非在__slots__
定义了__dict__
...
You can't create a type with a non-empty __slots__ attribute. 您无法创建具有非空__slots__属性的类型。 What you can do is insert a __slots__ attribute into the new class's dict, like this:
你可以做的是在新类的dict中插入__slots__属性,如下所示:
class Meta(type):
def __new__(cls, name, bases, dctn):
dctn['__slots__'] = ( 'x', )
return type.__new__(cls, name, bases, dctn)
class Foo(object):
__metaclass__ = Meta
def __init__(self):
pass
Now Foo has slotted attributes: 现在Foo有一些插槽属性:
foo = Foo()
foo.y = 1
throws 投
AttributeError: 'Foo' object has no attribute 'y'
dctn
in your example of a metaclass is the class dictionary, not the instance dictionary. 您的元类示例中的
dctn
是类字典,而不是实例字典。 __slots__
replaces the instance dictionary. __slots__
替换实例字典。 If you create two examples: 如果您创建两个示例:
class Meta(type):
def __new__(cls, name, bases, dctn):
return type.__new__(cls, name, bases, dctn)
class Foo1(object):
__metaclass__ = Meta
class Foo2(object):
__metaclass__ = Meta
__slots__ = ['a', 'b']
Then: 然后:
>>> f1 = Foo1()
>>> f2 = Foo2()
>>> f1.__dict__ is Foo1.__dict__
False
>>> f2.__dict__
Traceback (most recent call last):
...
AttributeError: 'Foo2' object has no attribute '__dict__'
if type
's subclass defines non-empty __slots__
, Python throws a TypeError
because of some complicated implementation-related stuff . 如果
type
的子类定义非空的__slots__
,Python会因为某些复杂的实现相关内容而抛出TypeError
。
In [1]:
class Meta(type):
__slots__ = ('x')
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-38-1b16edef8eca> in <module>()
----> 1 class Meta(type):
2 __slots__ = ('x')
TypeError: nonempty __slots__ not supported for subtype of 'type'
Empty __slots__
on the other hand don't produce any errors, but seam to have no effect. 另一方面,空
__slots__
不会产生任何错误,但接缝无效。
In [2]:
class Meta(type):
__slots__ = ()
class Foo(metaclass=Meta):
pass
type(Foo)
Out [2]:
__main__.Meta
In [3]:
Foo.y = 42
Foo.y
Out [3]:
42
In [4]:
Foo.__dict__
Out [4]:
mappingproxy({'y': 42, '__dict__': <attribute '__dict__' of 'Foo' objects>, '__doc__': None, '__weakref__': <attribute '__weakref__' of 'Foo' objects>, '__module__': '__main__'})
In [5]:
foo = Meta('foo', (), {})
type(foo).__slots__
Out [5]:
()
In [6]:
foo.x = 42
foo.x
Out [6]:
42
In [7]:
foo.__dict__
Out [7]:
mappingproxy({'__dict__': <attribute '__dict__' of 'foo' objects>, 'x': 42, '__module__': '__main__', '__doc__': None, '__weakref__': <attribute '__weakref__' of 'foo' objects>})
In [8]:
# Testing on non-metaclasses. Just in case.
class Bar:
__slots__ = ()
b = Bar()
b.__dict__
---------------------------------------------------------------------------
AttributeError Traceback (most recent call last)
<ipython-input-54-843730c66f3f> in <module>()
3
4 b = Bar()
----> 5 b.__dict__
AttributeError: 'Bar' object has no attribute '__dict__'
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