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
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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. | [reply]
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