元类和 prepare ()

Question

我正在自学 __prepare__ 函数。我在 PEP3115

看到了这个片段

# The custom dictionary
class member_table(dict):
    def __init__(self):
        self.member_names = []

    def __setitem__(self, key, value):
        # if the key is not already defined, add to the
        # list of keys.
        if key not in self:
            self.member_names.append(key)

        # Call superclass
        dict.__setitem__(self, key, value)

# The metaclass
class OrderedClass(type):

    # The prepare function
    @classmethod
    def __prepare__(metacls, name, bases): # No keywords in this case
        return member_table()

    # The metaclass invocation
    def __new__(cls, name, bases, classdict):
        # Note that we replace the classdict with a regular
        # dict before passing it to the superclass, so that we
        # don't continue to record member names after the class
        # has been created.
        result = type.__new__(cls, name, bases, dict(classdict))
        result.member_names = classdict.member_names
        return result

class MyClass(metaclass=OrderedClass):
    # method1 goes in array element 0
    def method1(self):
        pass

    # method2 goes in array element 1
    def method2(self):
        pass

我的问题是在这一行： result.member_names = classdict.member_names

变量 classdict 如何从 member_table class 中获取属性？我看到 __prepare__ 函数 returns 是 member_table 的实例，但是 member_table() 和 classdict.member_names 之间的 link 是如何生成的？

非常感谢大家！

Answer 1

这非常简单，因为这正是 prepare 所做的。

3.3.3.3. Preparing the class namespace Once the appropriate metaclass has been identified, then the class namespace is prepared. If the metaclass has a __prepare__ attribute, it is called as namespace = metaclass.__prepare__(name, bases, **kwds) (where the additional keyword arguments, if any, come from the class definition).

If the metaclass has no __prepare__ attribute, then the class namespace is initialised as an empty ordered mapping.

https://docs.python.org/3/reference/datamodel.html#preparing-the-class-namespace

这意味着，传递给 metaclass __new__ 和 __init__ 方法的 classdict 属性是完全相同的对象，即 return由 __prepare__ 编辑。

那个对象应该是一个映射实例，也就是说，一个行为像字典的对象并且至少有 __setitem__ 方法。此 __setitem__ 方法由 Python 为声明的 class 主体本身内设置的所有变量调用。

也就是说，对于没有自定义元class的普通class，变量记录在字典中（有序字典，自Python 3.6 起）。

当 Python 运行 class 主体内的每个语句时，就会发生这种情况。如果在 class 正文中调用 locals()，这就是 returned 的同一个对象：

In [21]: class M(type):
    ...:     @classmethod
    ...:     def __prepare__(cls, *args):
    ...:         class CustomDict(dict):
    ...:             __repr__ = lambda self: "I am a custom dict: " + str(id(self))
    ...:         namespace = CustomDict()
    ...:         print("From __prepare__", namespace)
    ...:         return namespace
    ...: 
    ...:     def __new__(metacls, name, bases, namespace):
    ...:         print("From __new__:", namespace)
    ...:         return super().__new__(metacls, name, bases, namespace)
    ...:     
    ...:     

In [22]: class Test(metaclass=M):
    ...:     def __init__(self):
    ...:         ...
    ...:     print("From class body:", locals(), locals()["__init__"])
    ...:     
    ...:     
From __prepare__ I am a custom dict: 140560887720440
From class body: I am a custom dict: 140560887720440 <function Test.__init__ at 0x7fd6e1bd7158>
From __new__: I am a custom dict: 140560887720440

最初设计此功能时的主要用例可能正是使 class 体内的声明顺序有意义的可能性。也就是说，__prepare__ 方法可以只是 return 一个 collections.OrderedDict 实例，而 __new__ 或 __init__ 将按照该顺序执行。从 Python 3.6 开始，class 属性的排序是默认的 - 而 __prepare__ 功能仍然非常先进，人们真的不得不考虑它的用途。

编辑： 通常 __prepare__ 作为 class 方法实现，因为它在 meta 之前被调用class 已实例化。

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