python - Base class method calling the Derived class method when using super. Very confused -

this seems confusing me. explain why unknown magical things happening ?

class a(object):     def testa(self):         print "testa of a"         self.testb()      def testb(self):         print "testb of a"  class b(a):     def testa(self):         super(b, self).testa()         print "testa of b"         self.testb()      def testb(self):         print "testb of b"  if __name__ == '__main__':     test = b()     test.testa() 

     program output:     ===============     testa of     testb of b --> why calling derived class method ?     testa of b     testb of b      expected output:     ================     testa of     testb of -- want see here.     testa of b     testb of b 

your answer appreciated. thank you.

in a.testa, call self.testb. means call "leaf" definition of testb current instance. since self instance of testb, calls b.testb.

even though wrote self.testb inside method defined on class a, not mean call version of method defined on class a. calling method on instance, , @ runtime class of instance determined dynamically, , whatever method defined on instance run. since instance of class b, , class b overrides testa, instance's version of testa 1 provided b.

if in a.testa want call a.testb, have explicitly calling a.testb(self). however, should think why want that. whole point of overriding methods can change how class things. a should not need know version of testb called. should need know calling method called testb whatever method testb documented in program/library. if a requires own testb method called, makes difficult subclasses alter behavior of testb, because a ignore overrides , keep calling own version instead.