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

by moritz (Cardinal)
on Feb 29, 2008 at 10:10 UTC ( [id://671136]=note: print w/replies, xml ) Need Help??


in reply to Changing local variables in subroutine by passing by reference?

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.

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Re^2: Changing local variables in subroutine by passing by reference?
by Anonymous Monk on Feb 29, 2008 at 10:27 UTC

    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?!

      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
      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,

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