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References: <ira8ti$6no$1@dont-email.me>
Date: Sat, 21 May 2011 23:27:31 -0700
Subject: Re: count strangeness
From: Chris Rebert <crebert@ucsd.edu>
To: James Stroud <jstroud@mbi.ucla.edu>
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On Sat, May 21, 2011 at 11:02 PM, James Stroud <jstroud@mbi.ucla.edu> wrote=
:
> tal 65% python2.7
> Python 2.7.1 (r271:86832, May 21 2011, 22:52:14)
> [GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin
> Type "help", "copyright", "credits" or "license" for more information.
> py> class C(object):
> ... =C2=A0 def __init__(self):
> ... =C2=A0 =C2=A0 self.data =3D []
> ... =C2=A0 def doit(self, count=3D0):
> ... =C2=A0 =C2=A0 for c in self.data:
> ... =C2=A0 =C2=A0 =C2=A0 count +=3D c.doit(count)
> ... =C2=A0 =C2=A0 count +=3D 1
> ... =C2=A0 =C2=A0 print count
> ... =C2=A0 =C2=A0 return count
> ...
> py> c =3D C()
> py> c.data.extend([C() for i in xrange(10)])
> py> c.doit()
> 1
> 2
> 4
> 8
> 16
> 32
> 64
> 128
> 256
> 512
> 1024
> 1024
>
> WTF?

Assuming your question is "Why is 1024 there twice?", the answer is
that it was the return value of the initial c.doit() call, and the
interactive interpreter outputs all non-None expression results; for
example:
[In the Python REPL]
>>> 5 + 5
10
>>>

So, you explicitly print the 1024 in your function once, and the
interpreter implicitly outputs it the second time. Notice what happens
when we use a statement instead of an expression:
>>> final =3D c.doit()
1
2
4
8
16
32
64
128
256
512
1024
>>> final
1024
>>>

And if we comment out the `print` in C.doit() and run your example again:
>>> c =3D C()
>>> c.data =3D [C() for i in xrange(10)]
>>> c.doit()
1024
>>>

Cheers,
Chris
--
http://rebertia.com