python - 具有 "super"属性的 Python 单继承中的 TypeError

Question

没想到Python2.7的继承如此反人类。为什么下面的代码给我一个 TypeError？

>>> class P:
...     def __init__(self, argp):
...         self.pstr=argp
... 
>>> class C(P):
...     def __init__(self, argp, argc):
...         super.__init__(self,argp)
...         self.cstr=argc
... 
>>> x=C("parent","child")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 3, in __init__
TypeError: descriptor '__init__' requires a 'super' object but received a 'instance'

然后我改成这2个，还是模糊的错误:

>>> class C(P):
...     def __init__(self, argp, argc):
...         super().__init__(self,argp)
...         self.cstr=argc
... 
>>> x=C("parent", "child")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 3, in __init__
TypeError: super() takes at least 1 argument (0 given)


>>> class C(P):
...     def __init__(self, argp, argc):
...         super(P).__init__(self,argp)
...         self.cstr=argc
... 
>>> x=C("parent", "child")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 3, in __init__
TypeError: must be type, not classobj

--------添加--------

还是不行，如下图

>>> class P:
...     def __init__(self, argp):
...         self.pstr=argp
... 
>>> class C(P):
...     def __init__(self, arg1, arg2):
...         super(C, self).__init__(arg1)
...         self.cstr=arg2
... 
>>> z=C("PPP", "CCC")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 3, in __init__
TypeError: must be type, not classobj

------------ADD2------------

似乎下面的 2 个代码片段都有效，有什么区别？顺便说一句，什么时候应该使用新样式类或旧样式类？

>>> class C(P):
...     def __init__(self, argp, argc):
...         super(C,self).__init__(argp)   #<------ C and self
...         self.cstr=argc
... 
>>> class C(P):
...     def __init__(self, argp, argc):
...         super(P,C).__init__(argp)  #<----- P and C
...         self.cstr=argc

Answer 1

根据您使用的 Python 版本，您的问题可能有两种回答。

python 2.x

super 必须这样调用 (example for the doc) :

class C(B):
    def method(self, arg):
        super(C, self).method(arg)

在您的代码中，这将是:

>>> class P(object):    # <- super works only with new-class style
...     def __init__(self, argp):
...         self.pstr=argp
... 
>>> class C(P):
...     def __init__(self, argp, argc):
...         super(C, self).__init__(argp)
...         self.cstr=argc

在这种情况下，您也不必在参数中传递 self。

python 3.x

在 Python 3.x 中，super 方法得到了改进:您可以 use it without any parameters (现在两者都是可选的)。此解决方案也可以工作，但仅适用于 Python 3 及更高版本:

>>> class C(P):
...     def __init__(self, argp, argc):
...         super().__init__(argp)
...         self.cstr=argc