Re: Unknown level of hash

On Mon, Mar 28, 2005 at 03:06:41PM -0800, Zhuang Li wrote:
> Hi, given an array: @a = ('E1', 'E2', ..., 'En'); 
> 
> Is there an easy way, hopefully one liner, to do the following without a
> loop? If not, will Perl support this in Perl 6?
> 
> $hash->{E1}->{E2}->...->{En} = 1;

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;
    }

(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.)

Andrew