Re: Parsing your script's command line

[Not_a_Number]

in reply to Parsing your script's command line

Very nice, ++

However, I have a small question on this snippet:

my $thing = shift;
$thing ||= 'default' unless defined $thing;

Surely the || is redundant here?

my $thing = shift;
$thing = 'default' unless defined $thing;

This seems to do the job just as well?

Or am I missing something?

thx

dave

[yosefm]

This is because $thing could be defined but zero and we don't want to overwrite a perfectly valid zero from the command line.

[Not_a_Number]

Er...

I'm afraid I don't understand. Under what circumstances does:

$thing = 'default' unless defined $thing;

not work for $thing == 0? I'm confused. Could you show me some code where the behaviour of the above differs from:

$thing ||= 'default' unless defined $thing;

Thanks in advance,

dave

[JadeNB]

This is because $thing could be defined but zero and we don't want to overwrite a perfectly valid zero from the command line.

As Not_a_Number points out, the syntax EXPR unless defined $thing will do nothing at all * if $thing is defined, whether it's true or false. That is, if we are executing EXPR, then $thing is guaranteed undefined, hence false; so $thing ||= 'default' is guaranteed to be the same as $thing = 'default'.

It seems reasonable to guess that what happened is that the coder originally had $thing ||= 'default' in some old code, discovered (as you mention) that it doesn't work when $thing is false-but-defined, and added the defined check without realising that it made ||= redundant.

Of course, a mere five years later, we have the wonderful //= (C style Logical Defined Or) instead to save us this pain.

UPDATE (the *'d statement above—sorry, I don't know how to do footnotes): On further thought, it's not quite true that EXPR unless defined $thing will do nothing if $thing is defined. Among what I suppose are many other subtle cases, if the unless is the last line in a subroutine, it'll make the subroutine return 1 when $thing is defined. For example, after

our $a = 1;
sub b { 0 unless defined $a }
sub c {}
my $b = b;
my $c = c;

we have that $b = 1 but $c is undefined.