The output is:
Class A
Class B
printout
Given code:
class A(object):
def __init__(self):
print "Class A"
def printout(self):
print "printout"
class B(A):
def __init__(self):
print "Class B"
def main():
myA = A()
myB = B()
myB.printout()
if __name__ == '__main__':
main()
I was hoping for:
Class A
Class A
Class B
printout
as result... :/
It's because you did not call the superclass's __init__.
class B(A):
def __init__(self):
A.__init__(self)
# ^^^^^^^^^^^^^^^^ this have to be explicitly added
print "Class B"
In Python 3.x you could use super().__init__() instead of A.__init__(self), but you're still explicitly invoking the superclass's __init__.
You need to explicitly call the base class constructor like so:
class B(A):
def __init__(self):
super(B, self).__init__()
In python 3.0 the following is equivilent:
class B(A):
def __init__(self):
super().__init__()
You didn't call the ancestor constructor, use the super() call for this
class B(A):
def __init__(self):
super(B, self).__init__()
print "Class B"
