Re: Unknown level of hash

	Hello,

Andrew Pimlott wrote:
 > If you're feeling functional,
 >
 >     ${fold_left(sub { \${$_[0]}->{$_[1]} }, \$hash, @a)} = 1;
 >
 > To get the value back out:
 >
 >     fold_left { $_[0]->{$_[1]}  } $hash, @a;
 >
 > Here is fold_left:
 >
 >     sub fold_left (&@) {
 >         my $sub = shift;
 >         my $acc = shift;
 >
 >         for my $each (@_) {
 >             $acc = $sub->($acc, $each);
 >         }
 >
 >         return $acc;
 >     }

	With your help, I found an almost-one-liner with a loop to do that :

use Data::Dumper;

@a = qw/a b c d/;
@b = qw/a 2 3 4/;
%hash = ();

# Preparation
$acc = \%hash;
$last = pop @a;

# Do the assignment
${ $acc = ($acc->{$_} ||= {}) for @a; \$acc }->{$last} = 1;

# Test on another array/value
$acc = \%hash;
$last = pop @b;
${ $acc = ($acc->{$_} ||= {}) for @b; \$acc }->{$last} = 2;

print Dumper(\%hash);

	But it could probably be written shorter.

> (You should be able to write the first one as
> 
>     ${fold_left { \${$_[0]}->{$_[1]} } \$hash, @a} = 1;
> 
> but Perl complains for no reason I can see.)

	Of course, the '{}' operator is a hash ref generator, not an anonymous 
sub declarator as 'sub {}' is...

	A+
-- 
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| |/ \|       Alexandre (Midnite) Jousset      |
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