用元类实现单例 [英] implement singleton with metaclass

问题描述

下面的代码是我如何使用元类实现单例，并且效果很好

the code below is how i implement singleton with metaclass, and it works well

class Test_MetaClass(type):

    def __init__(cls, name, bases, dict):
        super(Test_MetaClass, cls).__init__(cls, bases, dict)
        cls._instance = None
        print 'Test_MetaClass __init__'

    def __call__(cls, *args, **kwargs):
        if cls._instance is None:
            cls._instance = super(Test_MetaClass, cls).__call__(*args, **kwargs)
        print 'Test_MetaClass __call__'
        return cls._instance


class A(object):
    __metaclass__ = Test_MetaClass
    def __init__(self):
        print 'A __init__ triggered'

a = A()
b = A()

输出:

Test_MetaClass __init__
A __init__ triggered
Test_MetaClass __call__
Test_MetaClass __call__

我的问题是为什么b = A()直接进入Test_MetaClass.__call__而忽略__init__吗?

my question is why b = A() straight go to Test_MetaClass.__call__ and ignore __init__ ?

推荐答案

这不是您想要的吗? cls._instance不是None，因此不会执行type.__call__(cls, *args, **kwargs):

Isn't that what you wanted? cls._instance isn't None, so it won't execute type.__call__(cls, *args, **kwargs):

>>> type.__call__(A)
A __init__ triggered
<__main__.A object at 0x00BADB30>

通过此调用，调用A.__new__和A.__init__来创建/初始化新实例.但是对于您的单身人士，您只需要一个实例.

It's through this call that A.__new__ and A.__init__ get called to create/initialize a new instance. But for your singleton you only want one instance.

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