On 27/05/14 17:21, Abigail wrote:
On Tue, May 27, 2014 at 04:54:47PM +0100, Dirk Koopman wrote:
On 27/05/14 16:22, David Cantrell 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.
$bool = ($record->{perl} > 7) & 1; # for example?
$ perl -wE 'say +(8 > 7) & 1'
1
$
It would be very odd to consider 8 to be odd.
Indeed, hence my follow up.
But this works:
perl -e 'for (0..30) {print ((($_ > 7) && ($_ & 1)) ? "$_ = 1\n" : "$_ =
0\n")}';
Dirk
