Re: quick question about hash element access

Sorry, basically you have to do it recursively or using eval. For a miniscule efficiency gain you can use tail-optimization to turn the recursion into iteration like so:

sub nested_lookup {
  my $node = shift;
  while (ref($node) and @_) {
    $node = $node->{ shift(@_) };
  }
  return @_ ? undef : $node;
}
[download]

and you would call it like this:

my $elem = nested_lookup(\%hash, split /\./, 'a.b.c.d');
[download]

(Note: could benefit from more error checking.)

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Re: Re: quick question about hash element access by Chmrr (Vicar) on Jun 28, 2003 at 01:11 UTC
And with very few changes you can make it an lvalue, so you can use it to assign to: `sub nested : lvalue { @_ == 1 ? $_[0] : nested($_[0]{$_[1]}, @_[2..$#_]); }` [download] ..and now you can say: `nested(\%hash, a => b => c =>) = "VALUE";` TThis can lead to such fun looking constructs as `nested $hashref, a => b => c => d => = 42;` Note that this code will only work under perl 5.6.0 or later, though; earlier perls had more restricted definitions of lvalues. See also descending a tree of hash references. perl -pe '"I lo`+$^X$\"$]!$/"=~m%(.)%s;$_=$1;y^`+*^e v^#$&V"+@( NO CARRIER'	[reply] [d/l] [select]

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Re: Re: quick question about hash element access
by Chmrr (Vicar) on Jun 28, 2003 at 01:11 UTC

And with very few changes you can make it an lvalue, so you can use it to assign to:

sub nested : lvalue {
  @_ == 1
   ? $_[0]
   : nested($_[0]{$_[1]}, @_[2..$#_]);
}
[download]

..and now you can say:

nested(\%hash, a => b => c =>) = "VALUE";

TThis can lead to such fun looking constructs as nested $hashref, a => b => c => d => = 42; Note that this code will only work under perl 5.6.0 or later, though; earlier perls had more restricted definitions of lvalues. See also descending a tree of hash references.

perl -pe '"I lo*`+$^X$\"$]!$/"=~m%(.*)%s;$_=$1;y^`+*^e v^#$&V"+@( NO CARRIER'

[reply]
[d/l]
[select]