Arbitrary precision integer arithmetic: ceiling?
steve at REMOVE-THIS-cybersource.com.au
Sun Mar 9 03:38:18 CET 2008
On Sun, 09 Mar 2008 02:16:39 +0000, Steven D'Aprano wrote:
> On Sat, 08 Mar 2008 15:26:35 -0800, Paul Rubin wrote:
>> Alasdair <amca01 at gmail.com> writes:
>>> What is the best way of finding a ceiling of a quotient of arbitrary
>>> sized integers?
>> ceiling(a/b) = (a+b-1)//b
> Unfortunately that doesn't work reliably.
But of course you didn't intend it to work with floats, did you?
I'm just glad I didn't rant about people who think that floats are just
like reals when they're not.
More information about the Python-list