For those of you that are interested, I benchmarked a few of the
suggestions. Drieux is by a significant margin the winner speed-wise.
What I did was split the scalar into an array and then define the four
variables by join()ing the result using an array slice instead of making
a new array. Running this code, cutting out the print statements:
###################################################
use strict;
use warnings;
use Benchmark;
my $mine = timethese(1000000, {Mine => sub{
my $string = '0.0.0.0.1.10.1.30.1.10.1.30.1';
my @nums = split(/\./,$string);
my $first = join('.',@nums[0..3]);
my $second = $nums[4];
my $third = join('.',@nums[5..8]);
my $fourth = join('.',@nums[9..12]);
},Drieux => sub{
my $input = '0.0.0.0.111.10.1.30.1.10.1.30.1 ';
my $dot_quad = qr/\d+\.\d+\.\d+\.\d+/;
my ($im, $jm , $km, $mm ) =
$input =~
/($dot_quad)\.(\d+)\.($dot_quad)\.($dot_quad)/;
},Wags => sub{
$_ = '0.0.0.0.1.10.1.30.1.10.1.30.1';
my @MyWorka = ();
@MyWorka = split(/\./, $_);
my @MyWorkb = ();
my @MyWorkc = ();
my @MyWorkd = ();
@MyWorkb = splice(@MyWorka,-4,4);
@MyWorkc = splice(@MyWorka,-4,4);
@MyWorkd = splice(@MyWorka,-1,1);
}});
#####################################################
Gave me this result:
Benchmark: timing 1000000 iterations of Drieux, Mine, Wags...
Drieux: 9 wallclock secs ( 8.16 usr + 0.00 sys = 8.16 CPU) @
122518.99/s (n=1000000)
Mine: 12 wallclock secs (11.76 usr + 0.00 sys = 11.76 CPU) @
85055.71/s (n=1000000)
Wags: 14 wallclock secs (14.53 usr + 0.00 sys = 14.53 CPU) @
68818.39/s (n=1000000)
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