Re: incrementing values in base 36

Quoting Selector, Lev Y (Lev.Selector@xxxxxx):
> I have integer numbers represented in base 36 (each digit may have one of 36
> values: ('0'..'9','A'..'Z')).
> For example: '5BFHK'.
> I need to increment the number by 1 or by some step (for example, by 25).
> Here is my first take on incrementing by 1:

I don't have a fun way to do > 1 increments, but this is a fairly fun way to
do single increments:

@next{('0'..'9','A'..'Y', '')} = ('1'..'9','A'..'Z', '1');

sub advance {
    my($num) = @_;
    $num =~ s/([0-9A-Y]|)(Z*)$/$next{$1}.'0' x length($2)/e;
    return $num;
}
   
for (qw(5BFHK 111Z ZZZZ)) { print "\n$_\n",advance($_),"\n"; }

It exploits the fact that only Z does anything very interesting, and
interprets the strings as

AAANZZZZ

(where "A" is any alphanumeric, and "N" is not Z). The "N" is incremented,
the the "Z"s are all changed to "0". The mapping "" => "1" in %next takes
care of the case where the string is all Zs.

Any time I see Perl code looping through characters, I always try to turn it
into a regular expression. It's faster and often more readable.

I had a feeling when I was writing this that I should be using tr///
somehow, but I can't see how it would work. Maybe someone else can figure it
out.

Adam

-- 
Adam Rice -- wysiwyg@xxxxxxxxxxxxxxxxxxxxxx -- Blackburn, Lancashire, England