为什么在一个元类的call廑方法中使用super（）会引发TypeError？

Question

我使用的是python3，我发现我不能在元类的__call__内使用super()。在

为什么在下面的代码中super()引发TypeError: 'ListMetaclass' object is not iterable异常？为什么我从元类中删除__call__方法会很好地工作？在

class ListMetaclass(type):
    def __new__(cls, name, bases, attrs):
        attrs['add'] = lambda self, value: self.append(value)
        new_cls = type.__new__(cls, name, bases, attrs)
        return new_cls

    def __call__(cls, *args, **kw):
        ### why the 'super()' here raises TypeError: 'ListMetaclass' object is not iterable
        return super().__call__(cls, *args, **kw)
        return super(__class__, cls).__call__(cls, *args, **kw)
        return super(__class__, cls.__class__).__call__(cls, *args, **kw)

class MyList(list, metaclass=ListMetaclass):
    a=1
    def bar(self):
        print('test');

L=MyList()
L.add(1)
print('Print MyList :', L)

Answer 1

您不应该将cls传入super().__call__()；super()为您处理绑定，因此{}已经被自动传入了。在

您可能被__new__中的super().__new__(cls, ...)调用弄糊涂了，这是因为__new__是这里的例外，请参见^{} documentation：

Called to create a new instance of class cls. __new__() is a static method (special-cased so you need not declare it as such) that takes the class of which an instance was requested as its first argument.

从super().__call__(...)表达式中删除cls起作用：

class ListMetaclass(type):
    def __new__(cls, name, bases, attrs):
        attrs['add'] = lambda self, value: self.append(value)
        new_cls = type.__new__(cls, name, bases, attrs)
        return new_cls;
    def __call__(cls, *args, **kw):
        return super().__call__(*args, **kw)

通过传入cls，实际上是在执行list(cls)，告诉list()将{}转换为新列表中的值；这要求{}为iterable。在

当您移除元类上的__call__方法时，当您调用MyClass()时，将使用默认的type.__call__方法，该方法只接收常规参数（示例中没有）并返回一个新实例。在