Question

In Python, is there some short way to do something like

“for i in range(n)”

when n is too big for Python to actually create the array range(n)?

(short because otherwise I’d just use a while loop)

Answer 1

You could use xrange()… although that is restricted to “short” integers in CPython:

CPython implementation detail:
xrange() is intended to be simple and
fast. Implementations may impose
restrictions to achieve this. The C
implementation of Python restricts all
arguments to native C longs (“short”
Python integers), and also requires
that the number of elements fit in a
native C long. If a larger range is
needed, an alternate version can be
crafted using the itertools module:
takewhile(lambda x: x<stop,
(start+i*step for i in count())).

I don’t know whether that restriction also applies to other implementations (or which ones) – but there’s a workaround listed…

I know you mention bigint in your question title, but the question body talks about the number being too big to create the array – I suspect there are plenty of numbers which are small enough for xrange to work, but big enough to cause you memory headaches with range.