Catherine Heathcote wrote:
If I create a new class inherited from another with a constructor, what
happens with the new class's constructer?
assuming that you mean "is it called or not?":
Python doesn't really have constructors; when you create an object,
Python first creates the object and then calls the __init__ method, if
available
that method is an ordinary method, and behaves like all other methods:
if you override it, your version is used. if you don't, the other one
is used.
>>> class Parent:
... def __init__(self):
... print "init parent"
...
>>> class Child(Parent):
... pass
...
>>> o = Child()
init parent
>>> class Child(Parent):
... def __init__(self):
... print "init child"
...
>>> o = Child()
init child
if you want to override the parent's init method, but still use its
functionality, you can add an explicit call to the method:
>>> class Child(Parent):
... def __init__(self):
... Parent.__init__(self)
... print "init child"
...
>>> o = Child()
init parent
init child
since it's an ordinary method call, you don't have to initialize the
parent the first thing you do. you can also pass in any arguments you
want (including arguments passed to the child constructor):
>>> class OtherChild(OtherParent):
... def __init__(self, a, b, c):
... self.c = c
... OtherParent.__init__(self, a + b)
...
there's even a common pattern for passing along *all* arguments, no
matter what they are:
>>> class OtherChild(OtherParent):
... def __init__(self, *args, **kwargs):
... # do something here
... OtherParent.__init__(self, *args, **kwargs)
... # do something else here
...
instead of explicitly naming the baseclass, you can use the "super"
method to automatically look up the parent class, but this only works
if the parent class inherits from "object" or a subclass thereof:
>>> class Parent(object):
... def __init__(self):
... print "init parent"
...
>>> class Child(Parent):
... def __init__(self):
... super(Child, self).__init__()
... print "init child"
...
>>> o = Child()
init parent
init child
hope this helps!
</F>
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