Your problem is with the line

my $foo = shift || 1;

Consider what happens if you give this program a command line argument of 0.

"shift" returns the value of the first command line argument - which is 0, therefore "shift || 1" will be 1.

You need to think about the difference between the truth value of a variable and it's defined value. You want to check that your command line argument is defined - not that it is true.

You probably want something like this:

$foo = shift;
$foo = 1 unless defined $foo;

[download]

In Perl 5.10 you'll be able to use the "defined or" test to make this code easier.

$foo = shift // 1;

[download]
--
<http://www.dave.org.uk>

"The first rule of Perl club is you do not talk about Perl club."
-- Chip Salzenberg

In reply to Re: fun(?) with += by davorg
in thread fun(?) with += by Anonymous Monk

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