Re^2: Accessing Arguments inside Subroutines via @

I think what is really troubling you is the ambiguity between setting a symbol's slot and assigning to a symbol in Perl. (not sure if this is proper terminology)

Pure Perl has neither an explicit alias operator, nor an explicit unalias operator.˛

So if a scalar variable $a is an alias to $b you can't easily say $a=42 without changing $b , i.e. replacing the alias in the $a -"slot" with a literal 42.š

But when operating with arrays like @_ you have the choice.

$_[0]=42 is accessing the alias behind, but @_=(42) will replace the content of the slot.

  DB<106> sub repl { @_=(42); print "> $_[0]"; }

  DB<107> $a=666
 => 666

  DB<108> repl $a
> 42
  DB<109> $a
 => 666
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It has already been shown how to access multiple aliases at once (slicing), similarly you can use splice to access multiple slots of an array at once.

I hope the distinction between setting a slot and assigning to an alias is clearer now. :)

Cheers Rolf
_{(addicted to the Perl Programming Language and ☆☆☆☆ :)}

PS: Je suis Charlie!

š) But you can always use my $a=42 in a new scope or local $a=42 if it's a package var.

˛) like tie and untie

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Re^3: Accessing Arguments inside Subroutines via @_
by Athanasius (Archbishop) on Mar 22, 2015 at 09:04 UTC

Hello LanX,

I think what is really troubling you is the ambiguity between setting a symbol's slot and assigning to a symbol in Perl.

I’m not sure if I understand the distinction you’re making here, but my current understanding of a Perl alias is that it’s in some ways analogous to a smart pointer. That is, it has its own identity as a scalar value (of type “alias”), but under certain conditions it behaves “transparently” as the entity it aliases, like a dereferenced pointer. To explain what I mean: I was experimenting with various permutations of the following code:

Read more... (1162 Bytes)

and the output shows that following unshift @_ the element $_[1] is now an alias for the first argument. So, when explaining how @_ functions in a subroutine, I would no longer say “$_[0] is an alias for the first argument,” but rather “$_[0] is initialised to a (scalar value which is an) alias for the first argument.”

And the difference in behaviour between, e.g., @_ = reverse @_ and ( $_[0], $_[1] ) = ( $_[1], $_[0] ), comes down to whether the aliases act “transparently” or ”opaquely.” When elements of @_ are accessed individually — whether singly or as part of an array slice — they behave “transparently,” but when the array @_ is assigned-to — via = or splice — its alias values are accessed “opaquely,” meaning their referents remain unaffected.

Well, that’s my current understanding, and it may be just a convoluted way of repeating your explanation in different words. Anyway, I think I now understand what’s going on a little better than I did before. :-)

Update: Fixed typo.

Thanks for the help,

Athanasius <°(((>< contra mundum Iustus alius egestas vitae, eros Piratica,

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Re^4: Accessing Arguments inside Subroutines via @_

by LanX (Saint) on Mar 22, 2015 at 12:06 UTC

I took a look into the Panther book and experimented with Devel::Peek and the implementation of aliases seems simpler than a magic data structure of type alias.

Arrays (AVs) are composed from C-structures with pointers to C-structure(s) realizing scalars (SVs).

So $_[0]=666 updates the underlying scalar ( what I called "assigning" ) while @_=(666) creates the pointer to a new scalar ( "setting" ).

In the following dump the scalar value for $x is held within SV = IV(0x88f2498) at 0x88f249c

The array's first slot will point to that same structure before and after updating with $_[0]=666 (#markers added)

But creating new entries with @_=(666) will replace the pointer in this AV-slot to a new SV.

I hope the "how" is clearer now! :)

Cheers Rolf
_{(addicted to the Perl Programming Language and ☆☆☆☆ :)}

PS: Je suis Charlie!

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Re^3: Accessing Arguments inside Subroutines via @_
by AppleFritter (Vicar) on Mar 21, 2015 at 16:59 UTC

Pure Perl has neither an explicit alias operator

You can (ab)use typeglobs for this, at least for package variables. For instance:

#!/usr/bin/perl

no strict;
use warnings;
use feature qw/say/;

$a = 42;

*b = *a;
say $b; # 42

$b = 69;
say $a; # 69
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I can't think of a way of doing this with lexicals off the top of my head, though.

nor an explicit unalias operator.˛
So if a scalar variable $a is an alias to $b you can't easily say $a=42 without changing $b , i.e. replacing the alias in the $a -"slot" with a literal 42.š

I don't think this is needed. Suppose that you have variables - let's say lexicals - $a and $b, with the latter being aliased to the former. Unalias $a would mean creating a new lexical that's not related to $b anymore, so why not just declare a new lexical with the same value (my $c = $b) and use that instead? The result's the same, and it's arguably clearer, since if you look at any line in isolation there's no confusion over whether $a is (still) aliased to $b anymore.

OTOH there's something to be said in favor of explicit alias and unalias operators as well. I'd not be surprised at all if there were CPAN modules that implemented this.

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Re^4: Accessing Arguments inside Subroutines via @_

by LanX (Saint) on Mar 21, 2015 at 17:10 UTC

@_

my and local were already mentioned in my footnote, when using typeglobs I'd strongly suggest to prefer *b = \$a; over *b = *a;

But these mechanisms are no big help if you want named parameters in subs cause those are lexicals, (which should be the preferred variable flavor).

> I'd not be surprised at all if there were CPAN modules that implemented this.

see Data::Alias or Lexical::Alias and their discussion in PBP

Perl6 aims to handle aliasing consistently.

Cheers Rolf
_{(addicted to the Perl Programming Language and ☆☆☆☆ :)}

PS: Je suis Charlie!

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