Sorry for the length and code sloppiness...


Here's some sample code: 
use strict;
use warnings;

my $baz = 0;
my $foo = shift || 1;
chomp(my @bar = qw( etc etcetera ));

die "Nice Try." if $foo < 1;

while($baz < $foo){
$baz += 1;
print "$bar(bracket)rand @bar(/bracket) ", "ad infinitum\n";
}

[download]
Now, my problem with the code as it is(will change later in post) comes when I try to put 0 as a command line option when excecuting the script.

As it is, if you type 0 at the command line, it will still execute, with only one result, as though it had 1 or no option typed at the command line.

I'm guessing it has to do with the nature of the +=, that it will do something even when other commands tell it not to.

I did find a workaround, almost by accident.

Change || to or. However, if you use the default by not typing in an option, it will give this error:

Use of uninitialized value in numeric lt (<) at footest.pl line 9. Nice try. at footest.pl line 9.

Now, I tried a workaround, by changing "shift or 1," to "shift or die." This sort of works, but what happens is that, 1. You have to type a command line option or it will die, and more interestingly, 2. if you type 0 at the command line, it will give the error message for the "shift or die" line, rather than the other die message.

I just checked it by changing the second die line to < 2 and got the error messages like this:

perl footest.pl 1
Nice try. at footest.pl line 9.

perl footest.pl 0
Died at footest.pl line 6.

perl footest.pl -1
Nice try. at footest.pl line 9.

So...I'm thinking that 0 has some sort of special property here, that's making it act that way. Perhaps, as a guess, having to do with it's characteristic as being one of the things that means false?

20050102 Edit by ysth: code and p tags

In reply to fun(?) with += by Anonymous Monk

Title:
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