You will find some ideas in A better (ie.more concise) way to write this?.

Like this:

my $i = 0;
my $j = 1;
for( 1..10 ){
    ( $i+=1 ) %= 4;
    ( $j-=1 ) %= 4;
    printf "%d %d\n", $i+1, $j+1;
}
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Awesome, thanks guys, the code provided by kcott and hdb fit the bill for what I need. Thanks a lot!

One way is to use a closure to create a subroutine reference, this is a simpler version of anonymonk's &make_rollover.

$ perl -Mstrict -Mwarnings -E '
> my $cycle = do {
>    my $val = 0;
>    sub { return ( ( $val ++ ) % 4 ) + 1; };
>    };
> 
> say $cycle->() for 1 .. 10;'
1
2
3
4
1
2
3
4
1
2
$
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I hope this is useful.

Update: Corrected typo, s/ue/use/

range operator loops

my $whatever = 1;
while( $whatever ){
    for my $ix ( 1 .. 4 ){
        ...
    }
}
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modulus operator "cycle" through array

sub make_rollover {
    my( $completion_list ) = @_;
    my $ix = -1;
    my $xx = @$completion_list  - 1;
    sub {
        $ix++;
        return $completion_list->[  $ix % $xx ];
    };
}

my $rollover = make_rollover( \@tlud );
while ( my $next = $rollover->() ){
    ...
}
[download]

There may exist helper modules on cpan to embody these type of ... algorithm loops ...

Buy, sell, self, sleep, me :Drooooool

G'day Salamihawk,

Welcome to the monastery.

If @range holds your range of numbers and $i is initialised to zero, you can cycle through your numbers with:

$range[$i++ % @range]
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Here's my test:

#!/usr/bin/env perl -l

use strict;
use warnings;

my @range = 1 .. 4;
my $i = 0;

for (1 .. 9) {
    print $range[$i++ % @range];
}
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Output:

1
2
3
4
1
2
3
4
1
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my $i = 0;
# ... somewhere in the code:
$i = $i < 4 ? $i + 1 : 0;
[download]