Python 类本身是它们的 meta 类的对象 - 例如。

默认的金属类,当您确定类时应用于:

class foo:
    ...

例如,假设您正在构建一个ORM访问数据库,并且您希望每个表中的记录来自一个类地图到该表(基于字段,业务规则等),一个可能的使用MetaClass是例如,连接池逻辑,由所有表中的记录的所有类共享。

当你定义甲型时,你可以分类类型,并且可以超越下列魔法方法来插入你的逻辑。

class somemeta(type):
    __new__(mcs, name, bases, clsdict):
      """
  mcs: is the base metaclass, in this case type.
  name: name of the new class, as provided by the user.
  bases: tuple of base classes 
  clsdict: a dictionary containing all methods and attributes defined on class

  you must return a class object by invoking the __new__ constructor on the base metaclass. 
 ie: 
    return type.__call__(mcs, name, bases, clsdict).

  in the following case:

  class foo(baseclass):
        __metaclass__ = somemeta

  an_attr = 12

  def bar(self):
      ...

  @classmethod
  def foo(cls):
      ...

      arguments would be : ( somemeta, "foo", (baseclass, baseofbase,..., object), {"an_attr":12, "bar": <function>, "foo": <bound class method>}

      you can modify any of these values before passing on to type
      """
      return type.__call__(mcs, name, bases, clsdict)


    def __init__(self, name, bases, clsdict):
      """ 
      called after type has been created. unlike in standard classes, __init__ method cannot modify the instance (cls) - and should be used for class validaton.
      """
      pass


    def __prepare__():
        """
        returns a dict or something that can be used as a namespace.
        the type will then attach methods and attributes from class definition to it.

        call order :

        somemeta.__new__ ->  type.__new__ -> type.__init__ -> somemeta.__init__ 
        """
        return dict()

    def mymethod(cls):
        """ works like a classmethod, but for class objects. Also, my method will not be visible to instances of cls.
        """
        pass

无论如何,这两种是最常用的<unk>子,甲板是强大的,上面没有附近和完整的用途列表用于甲板。

在以对象为导向的编程中,一个代码类是一种类,其例子是类。 正如普通类定义某些对象的行为一样,一个代码类定义某些类的行为和其例子 这个术语的代码类简单地意味着用于创建类的东西。

什么是Metaclasses?你用它们用于什么?

>>> Class(...)
instance

>>> Metaclass(...)
Class

>>> type('Foo', (object,), {}) # requires a name, bases, and a namespace
<class '__main__.Foo'>

每当你创建一个类时,你都会使用一个类型:

class Foo(object): 
    'demo'

>>> Foo
<class '__main__.Foo'>
>>> isinstance(Foo, type), isinstance(Foo, object)
(True, True)

name = 'Foo'
bases = (object,)
namespace = {'__doc__': 'demo'}
Foo = type(name, bases, namespace)

>>> Foo.__dict__
dict_proxy({'__dict__': <attribute '__dict__' of 'Foo' objects>, 
'__module__': '__main__', '__weakref__': <attribute '__weakref__' 
of 'Foo' objects>, '__doc__': 'demo'})

(在 __dict__: __module__ 类的内容上有一个侧笔记,因为类必须知道它们在哪里定义,而 __dict__ 和 __weakref__ 是因为我们不定义 __slots__ - 如果我们定义 __slots__ 我们会在例子中节省一些空间,因为我们可以通过排除它们来排除 __dict__ 和 __weakref__。

>>> Baz = type('Bar', (object,), {'__doc__': 'demo', '__slots__': ()})
>>> Baz.__dict__
mappingproxy({'__doc__': 'demo', '__slots__': (), '__module__': '__main__'})

我们可以像任何其他类定义一样扩展类型:

>>> Foo
<class '__main__.Foo'>

class Type(type):
    def __repr__(cls):
        """
        >>> Baz
        Type('Baz', (Foo, Bar,), {'__module__': '__main__', '__doc__': None})
        >>> eval(repr(Baz))
        Type('Baz', (Foo, Bar,), {'__module__': '__main__', '__doc__': None})
        """
        metaname = type(cls).__name__
        name = cls.__name__
        parents = ', '.join(b.__name__ for b in cls.__bases__)
        if parents:
            parents += ','
        namespace = ', '.join(': '.join(
          (repr(k), repr(v) if not isinstance(v, type) else v.__name__))
               for k, v in cls.__dict__.items())
        return '{0}(\'{1}\', ({2}), {{{3}}})'.format(metaname, name, parents, namespace)
    def __eq__(cls, other):
        """
        >>> Baz == eval(repr(Baz))
        True            
        """
        return (cls.__name__, cls.__bases__, cls.__dict__) == (
                other.__name__, other.__bases__, other.__dict__)

>>> class Bar(object): pass
>>> Baz = Type('Baz', (Foo, Bar,), {'__module__': '__main__', '__doc__': None})
>>> Baz
Type('Baz', (Foo, Bar,), {'__module__': '__main__', '__doc__': None})

但是,与 eval(repr(Class))的进一步检查是不可能的(因为函数将是相当不可能从他们的默认 __repr__ 的 eval 。

from collections import OrderedDict

class OrderedType(Type):
    @classmethod
    def __prepare__(metacls, name, bases, **kwargs):
        return OrderedDict()
    def __new__(cls, name, bases, namespace, **kwargs):
        result = Type.__new__(cls, name, bases, dict(namespace))
        result.members = tuple(namespace)
        return result

class OrderedMethodsObject(object, metaclass=OrderedType):
    def method1(self): pass
    def method2(self): pass
    def method3(self): pass
    def method4(self): pass

>>> OrderedMethodsObject.members
('__module__', '__qualname__', 'method1', 'method2', 'method3', 'method4')

>>> inspect.getmro(OrderedType)
(<class '__main__.OrderedType'>, <class '__main__.Type'>, <class 'type'>, <class 'object'>)

而且它大约有正确的回报(除非我们能找到代表我们的功能的方式,否则我们就不能再评估):

>>> OrderedMethodsObject
OrderedType('OrderedMethodsObject', (object,), {'method1': <function OrderedMethodsObject.method1 at 0x0000000002DB01E0>, 'members': ('__module__', '__qualname__', 'method1', 'method2', 'method3', 'method4'), 'method3': <function OrderedMet
hodsObject.method3 at 0x0000000002DB02F0>, 'method2': <function OrderedMethodsObject.method2 at 0x0000000002DB0268>, '__module__': '__main__', '__weakref__': <attribute '__weakref__' of 'OrderedMethodsObject' objects>, '__doc__': None, '__d
ict__': <attribute '__dict__' of 'OrderedMethodsObject' objects>, 'method4': <function OrderedMethodsObject.method4 at 0x0000000002DB0378>})

看这:

Python 3.10.0rc2 (tags/v3.10.0rc2:839d789, Sep  7 2021, 18:51:45) [MSC v.1929 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> class Object:
...     pass
... 
>>> class Meta(type):
...     test = 'Worked!!!'
...     def __repr__(self):
...             return 'This is "Meta" metaclass'
... 
>>> class ObjectWithMetaClass(metaclass=Meta):
...     pass
... 
>>> Object or type(Object())
<class '__main__.Object'>
>>> ObjectWithMetaClass or type(ObjectWithMetaClass())
This is "Meta" metaclass
>>> Object.test
AttributeError: ...
>>> ObjectWithMetaClass.test
'Worked!!!'
>>> type(Object)
<class 'type'>
>>> type(ObjectWithMetaClass)
<class '__main__.Meta'>
>>> type(type(ObjectWithMetaClass))
<class 'type'>
>>> Object.__bases__
(<class 'object'>,)
>>> ObjectWithMetaClass.__bases__
(<class 'object'>,)
>>> type(ObjectWithMetaClass).__bases__
(<class 'type'>,)
>>> Object.__mro__
(<class '__main__.Object'>, <class 'object'>)
>>> ObjectWithMetaClass.__mro__
(This is "Meta" metaclass, <class 'object'>)
>>> 

换句话说,当一个对象没有创建(对象类型),我们正在寻找MetaClass。

请注意,在Python 3.6中,引入了一个新的Dunder方法 __init_subclass__(cls, **kwargs),以取代许多常见的使用案例为MetaClass。

型()函数可以返回对象的类型或创建一个新的类型,

例如,我们可以使用类()函数创建一个 Hi 类,并且不需要使用类 Hi(对象):

def func(self, name='mike'):
    print('Hi, %s.' % name)

Hi = type('Hi', (object,), dict(hi=func))
h = Hi()
h.hi()
Hi, mike.

type(Hi)
type

type(h)
__main__.Hi

除了使用类()以动态创建类,您还可以控制类的创建行为,并使用甲塔克拉斯。

根据 Python 对象模型,类是对象,所以类必须是另一个特定的类的例子. 默认情况下, Python 类是类类类的例子. 也就是说,类是大多数内置类的甲型类和用户定义类的甲型类。

class ListMetaclass(type):
    def __new__(cls, name, bases, attrs):
        attrs['add'] = lambda self, value: self.append(value)
        return type.__new__(cls, name, bases, attrs)

class CustomList(list, metaclass=ListMetaclass):
    pass

lst = CustomList()
lst.add('custom_list_1')
lst.add('custom_list_2')

lst
['custom_list_1', 'custom_list_2']

魔法将有效,当我们通过关键词论点在Metaclass,它指示Python翻译器通过ListMetaclass创建CustomList。新(),在此时,我们可以修改类定义,例如,并添加一个新的方法,然后返回修订的定义。