"Delaney, Timothy (Tim)" <[EMAIL PROTECTED]> writes:
> > Actually len(itertools.count()) would as well - when a couple of long
> > instances used up everything available - but it would take a *lot*
> > longer.
>
> Actually, this would depend on whether len(iterable) used a C integral
> variable to accumulate the length (which would roll over and never end)
> or a Python long (which would eventually use up all memory).
That's only because itertools.count itself uses a C int instead of a long.
IMO, that's a bug (maybe fixed in 2.5):
Python 2.3.4 (#1, Feb 2 2005, 12:11:53)
[GCC 3.4.2 20041017 (Red Hat 3.4.2-6.fc3)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import sys,itertools
>>> a=sys.maxint - 3
>>> a
2147483644
>>> b = itertools.count(a)
>>> [b.next() for i in range(8)]
[2147483644, 2147483645, 2147483646, 2147483647, -2147483648,
-2147483647, -2147483646, -2147483645]
>>>
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