Re: Assignment to a value only if it is defined

[ikegami]

my $bar = expensive();
$foo = $bar if defined $bar;

if (defined(my $bar = expensive())) {
   $foo = $bar;
}

$foo = $_ for grep defined, expensive();

sub assign_ifdef { $_[0] = $_[1] if defined($_[1]) }

assign_ifdef($foo, expensive());

The preceeding solution is very similar to what you asked (=ifdef) and very simple (single op).

Bonus points if $foo can also be a list.

I'm not sure if you mean

my $bar = expensive();
($i1, $i2) = $bar if defined $bar;

or

if (defined(my $bar = expensive()) {
   $_ = $bar for $i1, $i2;
}

Update: Added second and third.

[voj]

Thanks! That perfectly fits to assign to a scalar:

sub setifdef { $_[0] = $_[1] if defined($_[1]) }
setifdef $foo, $bar;

And in case of a list I'll use the (slightly less readable) 'for grep defined,' psuedo-inline operator. It only makes sense with list references (if you treat (undef) as valid list). Defined scalar values which are transformed to a list can also be handled:

@list = @{$_} for grep defined, $listref_or_undef;
@list = split(",",$_) for grep defined, $splitme_if_defined;