The next operator stops the current iteration of the loop and begins the next one. It does not assign $i to $j again. Modifying the loop control variable ($i = 0) and saying next creates an infinite loop. You can verify this by adding the following line as the first line of the loop:

print "I: $i\n";

Generally, loops like this are discouraged in Perl because there are better ways to do things. I might write your code like this: 

my $pos = 0;
for ($j .. $#array) {
    if ($array[$_] eq 'blah') {
        $pos = $_;
        last;
    }
}

[download]
By the end of my snippet, $pos will either contain the appropriate index, or will still be 0. It's shorter and needs no labels. Best of all, it doesn't need $i, so the question is moot.

I hope this helps.

Update: HyperZonk suggests that the original poster may want to search the first half of the array if $j through the end didn't have the element. In that case, maybe the C-style loop is better: 

my ($i, $end) = ($j, $#array);
my $pos = 0;
for (; $i <= $end; $i++) {
   last if ($array[$i] eq 'blah');
   if ($i == $#array) {
        $i = 0;
        $end = $j - 1;
   }
}

[download]
Untested, but it's one way to do it.

In reply to Re: Can I do this? by chromatic
in thread Can I do this? by Anonymous Monk

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