Re: How cand I find each dimension x,y,z of a 3D arrays

[jeroenes]

This code will recurse through the elements, and print the number of array-elements for each array in your structure. Will not decent hashes but can be hacked a bit to do that as well.

use strict;
 
my @a = ([([0..3] ) x 4]) x 3;
 
arraylength( \@a );
 
sub arraylength{
  my $a = shift;
  my $dim = 0;
  _length( $a, $dim );
}
 
sub _length{
  my $a = shift;
  my $dim = shift;
  if ( ref( $a ) =~ /ARRAY/ ){
    print "\t" x $dim;
    print "dim $dim -> ".(scalar @$a)." elements";
    print "\n\n" if $dim == 0;
    print "\n" if $dim == 1;
    _length( $_, $dim + 1) for ( @$a );
    print "\n";
  }
}

This example prints:

dim 0 -> 3 elements
 
        dim 1 -> 4 elements
                dim 2 -> 4 elements
                dim 2 -> 4 elements
                dim 2 -> 4 elements
                dim 2 -> 4 elements
 
        dim 1 -> 4 elements
                dim 2 -> 4 elements
                dim 2 -> 4 elements
                dim 2 -> 4 elements
                dim 2 -> 4 elements
 
        dim 1 -> 4 elements
                dim 2 -> 4 elements
                dim 2 -> 4 elements
                dim 2 -> 4 elements
                dim 2 -> 4 elements

[ton]

Very cool. But can I make one small suggestion? Use this function for length:

sub _length{
  my $a = shift;
  my $dim = shift;
  if ( ref( $a ) =~ /ARRAY/ ){
    print "\t" x $dim;
    print "dim $dim -> ".(scalar @$a)." elements";
    print "\n" if $dim == 0;
    print "\n";
    _length( $_, $dim + 1) for ( @$a );
  }
}

It's exactly the same logic, but prints newlines a little differently. Otherwise, you don't get the nice formatting for arrays with four or more dimensions.