Python using derived class's method in parent class?

Can I force a parent class to call a derived class's version of a function?

class Base(object):
    attr1 = ''
    attr2 = ''

    def virtual(self):
        pass               # doesn't do anything in the parent class

    def func(self):
        print "%s, %s" % (开发者_运维技巧self.attr1, self.attr2)
        self.virtual()

and a class that derives from it

class Derived(Base):
    attr1 = 'I am in class Derived'
    attr2 = 'blah blah'

    def virtual(self):
        # do stuff...
        # do stuff...

Clearing up vagueness:

d = Derived()
d.func()         # calls self.virtual() which is Base::virtual(), 
                 #  and I need it to be Derived::virtual()

If you instantiate a Derived (say d = Derived()), the .virtual that's called by d.func() is Derived.virtual. If there is no instance of Derived involved, then there's no suitable self for Derived.virtual and so of course it's impossible to call it.

It isn't impossible -- there is a way around this actually, and you don't have to pass in the function or anything like that. I am working on a project myself where this exact problem came up. Here is the solution:

class Base(): # no need to explicitly derive object for it to work
    attr1 = 'I am in class Base'
    attr2 = 'halb halb'

    def virtual(self):
        print "Base's Method"

    def func(self):
        print "%s, %s" % (self.attr1, self.attr2)
        self.virtual()

class Derived(Base):
    attr1 = 'I am in class Derived'
    attr2 = 'blah blah'

    def __init__(self):
  # only way I've found so far is to edit the dict like this
        Base.__dict__['_Base_virtual'] = self.virtual

    def virtual(self):
        print "Derived's Method"

if __name__ == '__main__':
    d = Derived()
    d.func()