Python和Singleton模式

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Python和Singleton模式

这个问题已经有了答案:

在Python中创建单例

似乎在Python中定义单例的方法有很多种。在Stack Overflow上是否有共识意见?

admin 更改状态以发布 2023年5月24日
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这是我自己实现的单例模式代码。你只需要在类上使用修饰符,然后通过调用Instance方法来获取该单例对象。以下是一个示例:

@Singleton
class Foo:
   def __init__(self):
       print 'Foo created'
f = Foo() # Error, this isn't how you get the instance of a singleton
f = Foo.instance() # Good. Being explicit is in line with the Python Zen
g = Foo.instance() # Returns already created instance
print f is g # True

这里是代码:

class Singleton:
    """
    A non-thread-safe helper class to ease implementing singletons.
    This should be used as a decorator -- not a metaclass -- to the
    class that should be a singleton.
    The decorated class can define one `__init__` function that
    takes only the `self` argument. Also, the decorated class cannot be
    inherited from. Other than that, there are no restrictions that apply
    to the decorated class.
    To get the singleton instance, use the `instance` method. Trying
    to use `__call__` will result in a `TypeError` being raised.
    """
    def __init__(self, decorated):
        self._decorated = decorated
    def instance(self):
        """
        Returns the singleton instance. Upon its first call, it creates a
        new instance of the decorated class and calls its `__init__` method.
        On all subsequent calls, the already created instance is returned.
        """
        try:
            return self._instance
        except AttributeError:
            self._instance = self._decorated()
            return self._instance
    def __call__(self):
        raise TypeError('Singletons must be accessed through `instance()`.')
    def __instancecheck__(self, inst):
        return isinstance(inst, self._decorated)

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我真的不认为有必要使用一个类来实现单例模式, 因为一个函数模块就足以充当单例。所有的变量都将绑定到该模块, 无论如何都不会重复实例化。

如果您确实想要使用类,那么在Python中无法创建私有类或私有构造函数,因此您无法对抗多次实例化,除非通过约定在使用API时进行保护。我仍然只会将方法放在一个模块中,并将该模块视为单例。

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