Why Perl boolean expression sometimes treated as lvalue?

vsespb has asked for the wisdom of the Perl Monks concerning the following question:

In continue of this http://www.perlmonks.org/?node_id=1017390 discussion

my $a = 1;
my $b = 2;
$a && $b = 3;
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this does not work. good. but this

use strict;
use warnings;
sub a
{
    $_[0] = 3;
}
my $x = 0;
my $y = 5;
a($x && $y);
print "$x\n";
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prints 3 (because $x aliased as 1-st arg to subrouting), so $x && $y works as lvalue. I wonder why so ? I didn't find documentation for this (not in perlsub/perlop)

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Re: Why Perl boolean expression sometimes treated as lvalue?
by Athanasius (Archbishop) on Feb 08, 2013 at 12:24 UTC

In Perl, arguments are passed by reference, not by value. This is what perlsub means when it says: “The array @_ is a local array, but its elements are aliases for the actual scalar parameters.”

The subroutine call a($x && $y) causes the Perl interpreter to evaluate the boolean expression $x && $y to determine which variable is to be aliased. If $x is false, the expression short-circuits, so a reference to $x is passed in to the sub; but if $x is true, short-circuiting does not occur, and so it is a reference to $y that is passed in to the sub.

So it is misleading to say that the boolean expression works as an l-value. Within sub a, the boolean expression is never seen — only the reference to either $x or $y — and it is this reference which works as an l-value.

I haven’t found anything in the documentation to explain the process by which Perl derives a reference (“alias”) from a complex expression given as an argument to a subroutine. It might be useful to know — but, in general, it’s safer and clearer to keep things simple and just pass a variable or a value.

Hope that helps,

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

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Re^2: Why Perl boolean expression sometimes treated as lvalue? (perlop, precedence)

by Anonymous Monk on Feb 08, 2013 at 12:37 UTC

perlop perlop perlop , precedence is everything :)

$ perl -E " print ( 1 && 2 )
2

$ perl -E " sub ff { ++$_[0] } my($f,$a)=(1,5); ff( $f && $a ); say $f
+; say $a; "
1
6
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Re: Why Perl boolean expression sometimes treated as lvalue?
by tobyink (Canon) on Feb 08, 2013 at 13:28 UTC

Hmmm... interesting disparity. The following may be of interest...

use v5.10;

my ($a, $b) = (0, 0);

sub lv :lvalue { $_[0] }

lv($a || $b) = 2;
lv($a && $b) = 1;

say $a;
say $b;
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package Cow { use Moo; has name => (is => 'lazy', default => sub { 'Mooington' }) } say Cow->new->name