Python 为什么是单身阶级'__调用了两次init_u;（）方法？

为什么

foo

打印两次

class A:
    _instance = None

    def __new__(cls, *args, **kwargs):
        if cls._instance is None:
            cls._instance = super().__new__(cls)
        return cls._instance

    def __init__(self, *, var):
        print('foo')
        self.var = var

a = A(var=1)
b = A(var=2)

assert a == b
assert a.var == b.var

当您编写

A（…）

时，您正在调用

type.__call__(A, ...)

这是因为

type

是

A

的元类<代码>类型.\uuuuuuuuuuuuu调用\uuuuuuu将依次调用

\uuuuuuuu新建\uuuuuuuuuu

和

\uuuuuuuuuuu初始化\uuuuuuuuuuu

<如果

\uuuu new\uuuu

返回类的实例，则将始终调用code>\uuuu init\uuuu。以下是一个简化视图：

def __call__(cls, *args, **kwargs):
    self = cls.__new__(cls, *args, **kwargs)
    if isinstance(self, cls):
        cls.__init__(self, *args, **kwargs)
    return self

我能想到的最简单的方法是将所有初始化逻辑放入

\uuuuu new\uuuu

：

class A:
    _instance = None

    def __new__(cls, *, var):
        if cls._instance is None:
            self = super().__new__(cls)
            self.var = var
            cls._instance = self

        return cls._instance

    def __init__(self, *args, **kwargs):
        print('foo')

当然，

foo

仍然会在每次请求新实例时打印出来，但它确实是一个单例

更全面的方法是重写元类调用方法的行为。这将避免调用

\uuuuu init\uuuuu

，除非调用一次：

class Singleton(type):
    def __call__(cls, *args, **kwargs):
        if hasattr(cls, '_instance'):
            return cls._instance
        return super().__call__(*args, **kwargs)

class A(metaclass=Singleton):
    def __init__(self, *, var):
        print('foo')
        self.var = var

---

因为每次调用

\uuuuu init\uuuuu

时，都会在

\uuuu new\uuuu

之后调用

。您使
\uuuuuuuuuuuuuuuuu
返回同一个对象的事实并不会阻止
\uuuuuuuuuuuuuuuuu
的执行。如果
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu>将返回现有对象，
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
应该是禁止操作的