>From 7 upwards is
('1' x $foo) =~ /^(11){3,}1$/
or something, I'm not even bothering to test this :S
On 27 May 2014 17:08, Jasper <jaspermcc...@gmail.com> wrote:
> Something like the prime regex would work:
>
> ('1' x $foo) =~ /^(11)*1$/
>
>
> On 27 May 2014 16:43, Joel Bernstein <j...@fysh.org> wrote:
>
>> Surely you only need to examine the right-most digit to know if the number
>> is odd? Your special requirement to (AIUI) consider 0..7 as even isn't
>> difficult to add.
>>
>> /joel
>>
>>
>> On 27 May 2014 16:22, David Cantrell <da...@cantrell.org.uk> wrote:
>>
>> > As part of the nasty mess that is CPANdeps, I have this line of code:
>> >
>> > $record->{is_dev_perl} = (
>> > $record->{perl} =~ /(^5\.(7|9|11|13|15|17|19|21)|rc|patch)/i
>> > ) ? 1 : 0;
>> >
>> > I'd like to not have to remember to add 23 to the list in a year or so's
>> > time. Can anyone think of a nice way of matching any odd number from 7
>> > upwards? Obviously it's easy to do in a coupla lines of perl code
>> > instead of a regex, so I'm asking more out of curiosity than because I
>> > actually need it.
>> >
>> > --
>> > David Cantrell | Official London Perl Mongers Bad Influence
>> >
>> > Today's previously unreported paraphilia is tomorrow's Internet
>> sensation
>> >
>> >
>>
>
>
>
> --
> Jasper
>
--
Jasper
