use strict;
use Data::Dumper;

my $ref = \&bar;
my @arry = foo($ref);
$ref = \&baz;
my $scal = foo($ref);

print Dumper @arry;
print Dumper $scal;

sub foo {
    my $ref = shift;
    my @a = &$ref;
    return wantarray ? @a : $a[0];
}

sub bar { (1..4) }
sub baz { [(1..4)] }
[download]

jeffa

That does not answer his question. If the return value of a function call is being stored in a list, the function is called in "array" context.

 @x  =  func()    ;  # list context
($x) =  func()    ;  # list context
 $x  = (func())[1];  # list context
[download]

Read my article on lists.

_____________________________________________________
Jeff[japhy]Pinyan: Perl, regex, and perl hacker.
s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;

Try it with:

$ref = sub {localtime(@_)};
[download]

You will very quickly find that your wrapper has changed rather dramatically what happens in scalar context.

Yep... this does work... I guess I just didn't fully understand how context stuff works...

                - Ant
                - Some of my best work - Fish Dinner