Question

My idea is any one of the objects should exist for all the subclasses of a Singleton class. The code that I have been trying and the result matrix is given below. The matrix seems to be working fine in the case of subclasses. Am I going the wrong way? Did it get what happens in the case of a Parent class object and a subclass object?

class Singleton(object):
    _instance = None
    def __new__(cls, *args, **kwargs):
        if not cls._instance:
            cls._instance = super(Singleton, cls).__new__(
                                cls, *args, **kwargs)
        return cls._instance

class A(Singleton):
    def __new__(cls, *args, **kwargs):
        super(A, cls).__new__(cls, *args, **kwargs)

class B(Singleton):
    def __new__(cls, *args, **kwargs):
        super(B, cls).__new__(cls, *args, **kwargs)

class C(B):
    def __new__(cls, *args, **kwargs):
        super(B, cls).__new__(cls, *args, **kwargs)


if __name__ ==  __main__ :
    s1=Singleton()
    s2=Singleton()
    if(id(s1)==id(s2)):
        print "Same"
    else:
        print "Different"

   
I got a result matrix for s1 and s2
           |************ s2 **************************|           
s1         |Singleton() |A()      | B()     | C()     |
===========|==========================================|
Singleton()|Same        |Different|Different|Different|
A()        |Different   |Same     |Same     |Same     |
B()        |Different   |Same     |Same     |Same     |
C()        |Different   |Same     |Same     |Same     |

Answer 1

<代码>new__> 方法有明确的<代码>return说明,因此,所有回归者均可获得<代码>None。因此,他们的新情况完全相同。这些代码与<代码>Singleton():

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

class A(Singleton):
    def __new__(cls, *args, **kwargs):
        return super(A, cls).__new__(cls, *args, **kwargs)

class B(Singleton):
    def __new__(cls, *args, **kwargs):
        return super(B, cls).__new__(cls, *args, **kwargs)

class C(B):
    def __new__(cls, *args, **kwargs):
        return super(C, cls).__new__(cls, *args, **kwargs)

甚至没有必要为次类确定<代码>new__():

class A(Singleton):
    pass

class B(Singleton):
    pass

class C(B):
    pass

Answer 2

法典看重的是罚款,但是,为某一类别及其子类收取单一吨值是不明智的。子类的全分点在某种程度上不同于母类和兄弟姐妹子类。我觉得奇怪的是,每一类人没有自己的单一州案例。

Answer 3

我在抽象课程的帮助下解决这一问题。在我看来:

from abc import ABCMeta, abstractmethod


class BasicClass(object):
    __metaclass__ = ABCMeta

    instance = None

    def __new__(cls, *args, **kwargs):
        if not cls.instance:
            cls.instance = super(BasicClass, cls).__new__(
                                    cls, *args, **kwargs)

        return cls.instance

    @abstractmethod
    def action(self):
        pass#do smthing


class FirstLevelChild(BasicClass):
    __metaclass__ = ABCMeta

    @abstractmethod
    def action(self):
        pass#do smthing in this or other abstract method
        # also pearent method can be called with help of super

class SecondLevelChild1(FirstLevelChild):
    def action(self):
        pass#do smthing unique for this class
        # in this or other abstract method
        # also pearent method can be called with help of super

class SecondLevelChild2(FirstLevelChild):
    def action(self):
        pass#do smthing unique for this class
        # in this or other abstract method
        # also pearent method can be called with help of super

检查

>>> a1 = SecondLevelChild1()
>>> a2 = SecondLevelChild1()
>>> b1 = SecondLevelChild2()
>>> b2 = SecondLevelChild2()
>>> a1 == b1
False
>>> a1 ==a2
True
>>> b1 == b2
True
>>>

ATTENTION! 如果你只需要上一级儿童班,这一解决办法将有所帮助。你们无法即时举办抽象课程。

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