So, the idiom is:

$random_item = $array[rand @array];
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Just cut-n-paste that!

-- Randal L. Schwartz, Perl hacker

My question wasn't clear. I know that rand() returns a number between 0 and it's argument. But my point was, it returns a decimal; i.e. 3.14159, 25.70098, .00001, etc. Does perl automatically round the number down(or take the integer portion)? For eaxample, here is my code:

    my $lRandomPassword = '';
    my $lMaxLength = 10;
    my $length;
    my $lRand;
    my @lKeySpace = ('A'..'Z', 'a..z', 0..9);
    my $lKeySpaceSize = scalar(@lKeySpace);
    for ($length = 0; $length < $lMaxLength; $length++) {
        $lRand = int(rand($lKeySpaceSize));
        $lRandomPassword .= $lKeySpace[$lRand];
    };
    $lRandomPassword;
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Yes, I know the code is overly verbose, but the important question for me is, could I safely get rid of the int() call? Does perl take the integer portion of an array index?

Yes it does automatically takethe integer portion when using it for an array index. Using int() you will also simply get the integer portion, not a rounded number, so you using the int() is sort of a noop. To round, you could do this:

my $foo = sprintf "%.0f", rand(100);
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But, to illustrate the answer to your question:

perl -wle '@r = (1,2,3); print $r[$t = rand(3)]; print "$t"';
2
1.28954588016495
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Cheers,
KM

Hmm. I'm not sure what you mean but I think you have made an incorrect assumption. Perl's int() just rounds down. It does not (necessarilly) convert the number into a long or unsigned long. You might want to read my previous node on Perl and floating point.

        - tye (but my friends call me "Tye")