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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 ( [id://671139]=note: print w/replies, xml ) Need Help??


in reply to Re: Changing local variables in subroutine by passing by reference?
in thread Changing local variables in subroutine by passing by reference?

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?!

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Re^3: Changing local variables in subroutine by passing by reference?
by ikegami (Patriarch) on Feb 29, 2008 at 11:12 UTC

    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
Re^3: Changing local variables in subroutine by passing by reference?
by moritz (Cardinal) on Feb 29, 2008 at 10:49 UTC
    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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