在python中，类型和类型的区别是什么？

我在写一个元类，不小心这样做了：

class MetaCls(type):
    def __new__(cls, name, bases, dict):
        return type(name, bases, dict)

class MetaCls(type):
    def __new__(cls, name, bases, dict):
        return type.__new__(cls, name, bases, dict)

MyClass = MyMeta.__new__(MyMeta, cls, bases, attributes)
meta_class = MyClass.__metaclass__
meta_class.__init__(MyClass, cls, bases, attributes)
return MyClass

…而不是像这样：

class MetaCls(type):
    def __new__(cls, name, bases, dict):
        return type(name, bases, dict)

class MetaCls(type):
    def __new__(cls, name, bases, dict):
        return type.__new__(cls, name, bases, dict)

MyClass = MyMeta.__new__(MyMeta, cls, bases, attributes)
meta_class = MyClass.__metaclass__
meta_class.__init__(MyClass, cls, bases, attributes)
return MyClass

---

这两个元类之间到底有什么区别？更具体地说，是什么导致第一个类不能正常工作（某些类没有被元类调用）？

描述得很好

如果没有返回正确类型的对象，那么定义自定义元类就没有意义

return type(name, bases, dict)

您从中得到的是一个新的

类型

，而不是一个

MetaCls

实例。因此，

MetaCls

（包括

\uuuu init\uuuu

）中定义的方法永远无法调用

---

type.\uuuuu new\uuuuu

将作为创建新类型的一部分被调用，是的，但是进入该函数的

cls

的值将是

type

而不是

MetaCls

，在第一个示例中，您正在创建一个全新的类：

>>> class MetaA(type):
...     def __new__(cls, name, bases, dct):
...         print 'MetaA.__new__'
...         return type(name, bases, dct)
...     def __init__(cls, name, bases, dct):
...         print 'MetaA.__init__'
... 
>>> class A(object):
...     __metaclass__ = MetaA
... 
MetaA.__new__
>>> 

在第二种情况下，您正在调用父级的

\uuuu new\uuuu

：

>>> class MetaA(type):
...     def __new__(cls, name, bases, dct):
...         print 'MetaA.__new__'
...         return type.__new__(cls, name, bases, dct)
...     def __init__(cls, name, bases, dct):
...         print 'MetaA.__init__'
... 
>>> class A(object):
...     __metaclass__ = MetaA
... 
MetaA.__new__
MetaA.__init__
>>> 

实现相同结果的另一种方法：

cls = "MyClass"
bases = ()
attributes = {'foo': 'bar'}
MyClass = MyMeta(cls, bases, attributes)

MyMeta

是可调用的，因此Python将使用特殊的方法

\uuuuuuu调用\uuuuuu

Python将在

MyMeta

的类型中查找

\uuuuuu调用

（在我们的例子中是

type

）

对于新样式的类，特殊方法的隐式调用是 只有在对象的类型上定义时才能保证正常工作，而不是 在对象的实例字典中“

MyClass=MyMeta（…）

被解释为：

my_meta_type = type(MyMeta)
MyClass = my_meta_type.__call__(MyMeta, cls, bases, attributes)

在

类型中。\uuu call\uuuu（）

我想象如下：

class MetaCls(type):
    def __new__(cls, name, bases, dict):
        return type(name, bases, dict)

class MetaCls(type):
    def __new__(cls, name, bases, dict):
        return type.__new__(cls, name, bases, dict)

MyClass = MyMeta.__new__(MyMeta, cls, bases, attributes)
meta_class = MyClass.__metaclass__
meta_class.__init__(MyClass, cls, bases, attributes)
return MyClass

MyMeta.\uuuuu new\uuuuuu（）

将决定如何构建

MyClass

：

type.\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

type（cls、base、attributes）
将为
MyClass
设置默认元类（即type），请参考下面的注释，希望这有帮助

class MetaCls(type):
    def __new__(cls, name, bases, dict):
        # return a new type named "name",this type has nothing
        # to do with MetaCls,and MetaCl.__init__ won't be invoked
        return type(name, bases, dict)

class MetaCls(type):
    def __new__(cls, name, bases, dict):
        # return a new type named "name",the returned type 
        # is an instance of cls,and cls here is "MetaCls", so 
        # the next step can invoke MetaCls.__init__ 
        return type.__new__(cls, name, bases, dict)

首先需要弄清楚
对象是如何工作的

这是下面的内容：

object.\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

调用以创建类
cls
的新实例<代码>\uuuu new\uuuuu（）是一个静态
采用
请求实例作为其第一个参数的类。
其余参数是传递给对象构造函数的参数
表达式（对类的调用）。
\uuuu new\uuuu（）的返回值
应该是新对象实例（通常是
cls
的实例）

典型的实现通过调用
超类的
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu（）方法使用
super（currentclass，cls）。\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
在返回新创建的实例之前，请根据需要将其删除

如果
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
已传递给
\uuuu new\uuuu（）

如果
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

\uuuu new\uuuu（）
主要用于允许不可变类型的子类（如
int
、
str
或
tuple
）自定义实例创建。也是
通常在自定义元类中重写以自定义类
创造

因此，在mg's中，前者不调用函数
\uuuu init\uuuuu
，而后者在调用
\uuu new\uuuuuu
后调用函数
\uuuu init\uuuuu
，但这就是我想问的问题：为什么
键入的对象不正确？我知道我必须使用子类的
\uuu new\uuu
方法，但是这与使用
类型有什么不同？类型如何神奇地知道你想要它生成什么类型的对象？因为我告诉了它。我输入了名称、基数和dict。我还需要提供哪些其他信息？类型是如何与对象相关的？我没有得到那部分。OP问的是关于
类型的问题。它创建了一个类，而不是
对象。它创建了一个类的实例。请看关于类类型（对象）的文档，我想这可能是类型和对象之间的关系。如果你明白了，记得取消你的否决票，谢谢！您（@Nishant，@rlat）需要弄清楚的是Python的类型和对象是什么，下面是答案；因此类型之间没有区别。与对象相关。对于python3
类A（对象，元类=MetaA）：pass
。class A作用域中的
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu。但这里你指的是类名。