http://www.perlmonks.org?node_id=283951


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

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Re: Re: Parsing your script's command line
by yosefm (Friar) on Sep 04, 2003 at 11:48 UTC
    This is because $thing could be defined but zero and we don't want to overwrite a perfectly valid zero from the command line.

      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

      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.