python: Constructor goes into infinite loop

Question

My purpose is to override some of the functions of 'First' class run-time for certain cases. So I want to derive a class from the original one. Here is the code snippet.

class First(object):

    def __init__(self):
        print "First"
        super(First, self).__init__()

    def foo(self):
        print "foo"

class Second(First):

    def __init__(self):
        print "second"
        super(Second, self).__init__()

    def foo(self):
        print "want to override this"

First = Second

o = First()

Why the constructor goes into infinite loop? What wrong thing am doing?

tag your programming language

Arun A S
– Arun A S

2015-03-02 10:19:27 +00:00
Commented Mar 2, 2015 at 10:19 — Arun A S
– Arun A S, Commented Mar 2, 2015 at 10:19

dhke · Accepted Answer · 2015-03-02 12:43:52Z

2

Python names are not looked up at compile time, name lookups happen when the code is executed.

The thing to watch out for is

First = Second

Because of the assignment, First() will create an instance of class Second
Second.__init__() will call First.__init__().
in First.__init__(), First will be looked up by name in the global context.
Since you reassigned First = Second, the name First points to class Second. Which will get its __init__() called and that gives you your infinite recursion.

In short: Don't do this ...

answered Mar 2, 2015 at 12:43

dhke

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2 Comments

ChandiG Over a year ago

instead of super() if I call parent constructor directly, it does not go into recursive loop.. however my purpose is to patch the original function runtime and not the class itself. Monkey patching kind of approach might help.

dhke Over a year ago

Helper tools for function/method wrapping and monkey patching are in functools, functools.wraps is your friend.

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