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python怎么实现单例模式

来源:中华考试网  [2020年11月6日]  【

  Python单例模式的4种实现方法:

  #-*- encoding=utf-8 -*-

  print '----------------------方法1--------------------------'

  #方法1,实现__new__方法

  #并在将一个类的实例绑定到类变量_instance上,

  #如果cls._instance为None说明该类还没有实例化过,实例化该类,并返回

  #如果cls._instance不为None,直接返回cls._instance

  class Singleton(object):

  def __new__(cls, *args, **kw):

  if not hasattr(cls, '_instance'):

  orig = super(Singleton, cls)

  cls._instance = orig.__new__(cls, *args, **kw)

  return cls._instance

  class MyClass(Singleton):

  a = 1

  one = MyClass()

  two = MyClass()

  two.a = 3

  print one.a

  #3

  #one和two完全相同,可以用id(), ==, is检测

  print id(one)

  #29097904

  print id(two)

  #29097904

  print one == two

  #True

  print one is two

  #True

  print '----------------------方法2--------------------------'

  #方法2,共享属性;所谓单例就是所有引用(实例、对象)拥有相同的状态(属性)和行为(方法)

  #同一个类的所有实例天然拥有相同的行为(方法),

  #只需要保证同一个类的所有实例具有相同的状态(属性)即可

  #所有实例共享属性的最简单最直接的方法就是__dict__属性指向(引用)同一个字典(dict)

  #可参看:http://code.activestate.com/recipes/66531/

  class Borg(object):

  _state = {}

  def __new__(cls, *args, **kw):

  ob = super(Borg, cls).__new__(cls, *args, **kw)

  ob.__dict__ = cls._state

  return ob

  class MyClass2(Borg):

  a = 1

  one = MyClass2()

  two = MyClass2()

  #one和two是两个不同的对象,id, ==, is对比结果可看出

  two.a = 3

  print one.a

  #3

  print id(one)

  #28873680

  print id(two)

  #28873712

  print one == two

  #False

  print one is two

  #False

  #但是one和two具有相同的(同一个__dict__属性),见:

  print id(one.__dict__)

  #30104000

  print id(two.__dict__)

  #30104000

  print '----------------------方法3--------------------------'

  #方法3:本质上是方法1的升级(或者说高级)版

  #使用__metaclass__(元类)的高级python用法

  class Singleton2(type):

  def __init__(cls, name, bases, dict):

  super(Singleton2, cls).__init__(name, bases, dict)

  Python单例模式的4种实现方法:  #-*- encoding=utf-8 -*-   print '----------------------方法1--------------------------'   #方法1,实现__new__方法  #并在将一个类的实例绑定到类变量_instance上,   #如果cls._instance为None说明该类还没有实例化过,实例化该类,并返回  #如果cls._instance不为None,直接返回cls._instance   class Singleton(object):   def __new__(cls, *args, **kw):   if not hasattr(cls, '_instance'):   orig = super(Singleton, cls)   cls._instance = orig.__new__(cls, *args, **kw)   return cls._instance   class MyClass(Singleton):   a = 1   one = MyClass()   two = MyClass()   two.a = 3   print one.a   #3   #one和two完全相同,可以用id(), ==, is检测  print id(one)   #29097904   print id(two)   #29097904   print one == two   #True   print one is two   #True   print '----------------------方法2--------------------------'   #方法2,共享属性;所谓单例就是所有引用(实例、对象)拥有相同的状态(属性)和行为(方法)   #同一个类的所有实例天然拥有相同的行为(方法),   #只需要保证同一个类的所有实例具有相同的状态(属性)即可  #所有实例共享属性的最简单最直接的方法就是__dict__属性指向(引用)同一个字典(dict)   #可参看:http://code.activestate.com/recipes/66531/   class Borg(object):   _state = {}   def __new__(cls, *args, **kw):   ob = super(Borg, cls).__new__(cls, *args, **kw)   ob.__dict__ = cls._state   return ob   class MyClass2(Borg):   a = 1   one = MyClass2()   two = MyClass2()   #one和two是两个不同的对象,id, ==, is对比结果可看出  two.a = 3   print one.a   #3   print id(one)   #28873680   print id(two)   #28873712   print one == two   #False   print one is two   #False   #但是one和two具有相同的(同一个__dict__属性),见:   print id(one.__dict__)   #30104000   print id(two.__dict__)   #30104000   print '----------------------方法3--------------------------'   #方法3:本质上是方法1的升级(或者说高级)版  #使用__metaclass__(元类)的高级python用法  class Singleton2(type):   def __init__(cls, name, bases, dict):   super(Singleton2, cls).__init__(name, bases, dict)   cls._instance = None   def __call__(cls, *args, **kw):   if cls._instance is None:   cls._instance = super(Singleton2, cls).__call__(*args, **kw)   return cls._instance   class MyClass3(object):   __metaclass__ = Singleton2   one = MyClass3()   two = MyClass3()   two.a = 3   print one.a   #3   print id(one)   #31495472   print id(two)   #31495472   print one == two   #True   print one is two   #True   print '----------------------方法4--------------------------'   #方法4:也是方法1的升级(高级)版本,   #使用装饰器(decorator),   #这是一种更pythonic,更elegant的方法,   #单例类本身根本不知道自己是单例的,因为他本身(自己的代码)并不是单例的  def singleton(cls, *args, **kw):   instances = {}   def _singleton():   if cls not in instances:   instances[cls] = cls(*args, **kw)   return instances[cls]   return _singleton   @singleton   class MyClass4(object):   a = 1   def __init__(self, x=0):   self.x = x   one = MyClass4()   two = MyClass4()   two.a = 3   print one.a   #3   print id(one)   #29660784   print id(two)   #29660784   print one == two   #True   print one is two   #True   one.x = 1   print one.x   #1   print two.x   #1

  cls._instance = None

  def __call__(cls, *args, **kw):

  if cls._instance is None:

  cls._instance = super(Singleton2, cls).__call__(*args, **kw)

  return cls._instance

  class MyClass3(object):

  __metaclass__ = Singleton2

  one = MyClass3()

  two = MyClass3()

  two.a = 3

  print one.a

  #3

  print id(one)

  #31495472

  print id(two)

  #31495472

  print one == two

  #True

  print one is two

  #True

  print '----------------------方法4--------------------------'

  #方法4:也是方法1的升级(高级)版本,

  #使用装饰器(decorator),

  #这是一种更pythonic,更elegant的方法,

  #单例类本身根本不知道自己是单例的,因为他本身(自己的代码)并不是单例的

  def singleton(cls, *args, **kw):

  instances = {}

  def _singleton():

  if cls not in instances:

  instances[cls] = cls(*args, **kw)

  return instances[cls]

  return _singleton

  @singleton

  class MyClass4(object):

  a = 1

  def __init__(self, x=0):

  self.x = x

  one = MyClass4()

  two = MyClass4()

  two.a = 3

  print one.a

  #3

  print id(one)

  #29660784

  print id(two)

  #29660784

  print one == two

  #True

  print one is two

  #True

  one.x = 1

  print one.x

  #1

  print two.x

  #1

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