python - 切换祖先顺序会破坏初始化吗？

Question

这是来自 Django 应用程序，以自定义形式使用多重继承:

from django import forms

class CommonForm(forms.ModelForm):
    class Meta:
        model = MyModel

    def __init__(self, request, *args, **kwargs):
        print 'CommonForm.__init__'
        super(CommonForm, self).__init__(request, *args, **kwargs)
        print 'CommonForm.__init__ FINISHED'
        self._request = request

class MyMixin(object):
    def __init__(self, *args, **kwargs):
        print 'MyMixin.__init__'
        super(MyMixin, self).__init__(*args, **kwargs)
        print 'MyMixin.__init__ FINISHED'

class CustomForm(CommonForm, MyMixin):
    def __init__(self, request, *args, **kwargs):
        print 'CustomForm.__init__'
        super(CustomForm, self).__init__(request, *args, **kwargs)
        print 'CustomForm.__init__ FINISHED'

# weird initialization:
frm = CustomForm(http_request)

# outputs:
CustomForm.__init__
CommonForm.__init__
CommonForm.__init__ FINISHED
CustomForm.__init__ FINISHED

什么会导致 MyMixin.__init__ 被跳过？! 没有错误消息，没有抛出异常，什么都没有......

如果我将类继承顺序切换为

class CustomForm(MyMixin, CommonForm):

所有初始化器都按预期突然调用:

frm = CustomForm(http_request)

CustomForm.__init__
MyMixin.__init__
CommonForm.__init__
CommonForm.__init__ FINISHED
MyMixin.__init__ FINISHED
CustomForm.__init__ FINISHED

谁能解释一下这是怎么回事吗？

Answer 1

我不确定为什么会发生这种情况，但我确实知道在使用多重继承时不应该使用 super()，而是应该:

    CommonForm.__init__(request, *args, **kwargs)
    MyMixin.__init__(*args, **kwargs)

这是一个很好的例子:http://www.python-course.eu/python3_multiple_inheritance.php