multiple inheritance execution order

Question

code1：

class base(object):
    def test(self):
        pass


class low1(object):
    def test(self):
        super(low1, self).test()
        print "low1 test"


class low2(object):
    def test(self):
        super(low2, self).test()
        print "low2 test"


class high(low1, low2, base):
    pass


if __name__ == "__main__":
    high().test()

code2：

class base(object):
    def test(self):
        pass


class low1(object):
    def test(self):
        # super(low1, self).test()
        print "low1 test"


class low2(object):
    def test(self):
        # super(low2, self).test()
        print "low2 test"


class high(low1, low2, base):
    pass


if __name__ == "__main__":
    high().test()

the output of code1 is：

low2 test
low1 test

the output of code2 is：

low1 test

when I call why test method of a high object， it execute test method of both low1 and low2？

You called `super(low2...` in code1. If you are going to use `super` please [learn to understand how they work and why you might use them](http://stackoverflow.com/questions/576169/understanding-python-super-with-init-methods). — metatoaster, Oct 30 '15 at 03:39
I just added the "Python 2" tag to the question. The reason is that the behaviour of inheritance and `super()` changed a bit between Python 2 and Python 3. BTW: If you're learning Python, don't start with Python 2 but use the current version 3 from the start, it offers quite a few advantages. — Ulrich Eckhardt, Oct 30 '15 at 06:48

score 2 · Accepted Answer · answered Oct 30 '15 at 08:52

Have a look at the method resolution order:

print(high.mro())

This prints:

[<class '__main__.high'>, <class '__main__.low1'>, <class '__main__.low2'>,
 <class '__main__.base'>, <class 'object'>]

Think of super() meaning "next in line", where line is the list of classes shown above. Therefore, super(low1, self) finds low2 as the class next in line.

multiple inheritance execution order

1 Answers1