Python 为什么“超级”.\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu？

试着去理解超新__

下面是我的代码：

class Base(object):
    def __new__(cls,foo):
        if cls is Base:
            if foo == 1:
                #  return Base.__new__(Child) complains not enough arguments
                return Base.__new__(Child,foo)
            if foo == 2:
                # how does this work without giving foo?
                return super(Base,cls).__new__(Child)  
        else:
            return super(Base,cls).__new__(cls,foo)

    def __init__(self,foo):
        pass 
class Child(Base):
    def __init__(self,foo):
        Base.__init__(self,foo)    
a = Base(1)  # returns instance of class Child
b = Base(2)  # returns instance of class Child
c = Base(3)  # returns instance of class Base
d = Child(1)  # returns instance of class Child

为什么“超级”.\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

Python:2.7.11

super.\uuuuuu new\uuuuuu与Base的函数不同。超级。新的是对象。新的。对象.\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

---

这将调用对象。\uuuu new\uuuu cls，foo。没错，它将一个foo参数传递给object.\uuuuu new\uuuu，即使object.\uuuuu new\uuu不需要它。这在Python2中是允许的，但在Python3中会崩溃。最好从那里删除foo参数。

原因与object相同。\uuuuu new\uuuuuu不需要foo而是Base。\uuu new\uuuuu需要。还有一个问题是，当子类定义在基之下时，我们如何在返回superBase cls时引用子对象。\uu new\uu Child？我的意思是，理想情况下，基地不应该在它的范围内有孩子，我想？那是因为超级拥有超级力量吗？@AnveshYalamarthy不，那不需要超级力量。这与为什么两个函数可以互相调用的原因相同，即使一个函数是在另一个函数之后定义的——在执行代码时，子类已经创建。如果在定义子类之前尝试a=Base1，那么确实会出现错误。

>>> Base.__new__
<function Base.__new__ at 0x000002243340A730>
>>> super(Base, Base).__new__
<built-in method __new__ of type object at 0x00007FF87AD89EC0>
>>> object.__new__
<built-in method __new__ of type object at 0x00007FF87AD89EC0>

return super(Base,cls).__new__(cls, foo)