RE: Simple Recursive Generator Question
From: Robert Brewer (fumanchu_at_amor.org)
Date: 12/19/03
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Date: Fri, 19 Dec 2003 11:50:27 -0800 To: <python-list@python.org>
Probably not what you meant, but one of the things you're missing is
that generators nullify the need for recursion in this case:
def big(mask):
index = 0
while True:
if mask == 0:
break
if mask & 0x1:
yield index
index += 1
mask = mask >> 1
>>> for x in big(0x16):
... print x
...
1
2
4
The generator saves you having to pass the "index" param recursively.
You save both object memory size (only one 'index' and one 'mask' object
and their bound names) and function-calling overhead (stack space).
Robert Brewer
MIS
Amor Ministries
fumanchu@amor.org
> -----Original Message-----
> From: MetalOne [mailto:jcb@iteris.com]
> Sent: Friday, December 19, 2003 11:14 AM
> To: python-list@python.org
> Subject: Simple Recursive Generator Question
>
>
> I am trying to write a generator function that yields the
> index position
> of each set bit in a mask.
> e.g.
> for x in bitIndexGenerator(0x16): #10110
> print x
> --> 1 2 4
>
>
> This is what I have, but it does not work.
> Changing yield to print, shows that the recursion works correctly.
>
> def bitIndexGenerator(mask, index=0):
> if mask == 0: return
> elif mask & 0x1: yield index
> bitIndexGenerator(mask >> 1, index+1)
>
> What am I missing?
> --
> http://mail.python.org/mailman/listinfo/python-list
>
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