Re: Changing local variables in subroutine by passing by reference?

[moritz]

You need to dereference your values before assigment:

sub my_sub {
   my $scalar_ref = $_[0];
   # note the $$: 
   $$scalar_ref = 'new value';
}
# and call it:
my_sub(\$variable);

# another way, since @_ is an alias:
sub my_sub_2 {
    $_[0] = 3;
}
my $a = 1;
my_sub_2($a);
print $a; # prints 3

But generally it is better style to modify your values on the caller side, based on the return value of the function.

[Anonymous Monk]

Thanks, I think that's working now :)

However, I don't think I understand the difference between "# another way, since @_ is an alias:" and what I was doing.. Is it because by assigning a reference to $_[0] (say) $_[0] would then be a reference to a reference? And if so, would my code have worked if I'd added an extra dereferencer (e.g. $$$_[0] instead of $$_[0]), or is that just nonsense?!

[ikegami]

I don't think I understand the difference between "# another way, since @_ is an alias:"

In Perl, everything is passed by reference. To avoid confusion with Perl references — an unrelated concept — this is known as "aliasing" in the Perl world.

When a function is called, the elements of @_ are aliased to the arguments in the caller. Changes to @_ will be reflected in the arguments passed to the function.

In the following snippet, $_[0] is aliased to $x, and $_[1] is aliased to $y. Notice how changing @_ changed $x and $y, and notice how no references — and no dereferencing — were used.

sub func {
   print("f-pre:  $_[0], $_[1]\n");
   $_[0] = 5;
   $_[1] = 6;
   print("f-post: $_[0], $_[1]\n");
}

my $x = 3;
my $y = 4;
print("pre:    $x, $y\n");
func($x, $y);
print("post:   $x, $y\n");

pre:    3, 4
f-pre:  3, 4
f-post: 5, 6
post:   5, 6

[moritz]

My second example didn't uses references at all! That's the differences.

Note that normally your subs are built like this:

sub my_sub {
   my ($arg1, $arg2) = @_;
   ...
}

The assignment to another variable breaks the alias, so if you change $arg1, @_ will not be modified. But if you modify @_ directly,