Re: Parse::RecDescent for simple syntax-directed translation

tomazos,
(This is the "classic example" of something that can't be done with regular expressions, and needs a context-free grammar. This is because within the regexp /(a*)(b*)/ there is no way to assert that length($1) == length ($2). Anyhoo...)

While this is probably strictly true, Perl is all about letting you get the job done:

my $str = "eeeeaaaabbbeeee";
$str =~ s/((a+)(??{'b'x length$2}))/'c' x (length($1) * .5) . 'd' x (l
+ength($1) * .5)/e;
[download]

Anyhoo...I have just recently started learning Parse::RecDescent. Update: I am not sure if this is what you had in mind, but the following accomplishes what you want without using sneaky experimental regex features.

#!/usr/bin/perl
use strict;
use warnings;
use Parse::RecDescent;

$Parse::RecDescent::skip = '';

my $grammar = q{
    match  : PREFIX TOKEN SUFFIX
             {print join '', @item[1..3]}
    PREFIX : /.*?(?=a+b+)/
    TOKEN  : /a+b+/
             {
                 my $str   = $item[1];
                 my $a_cnt = $str =~ tr/a//;
                 my $b_cnt = $str =~ tr/b//;
                 if ($a_cnt == $b_cnt) {
                     $return = ('c' x $a_cnt) . ('d' x $b_cnt);
                 }
                 elsif ($a_cnt > $b_cnt) {
                     $return = ('a' x ($a_cnt - $b_cnt)) . ('c' x $b_c
+nt) . ('d' x $b_cnt);
                 }
                 else {
                     $return = ('c' x $a_cnt) . ('d' x $a_cnt) . ('b' 
+x ($b_cnt - $a_cnt));
                 }
             }
    SUFFIX : /.*$/
};
my $parser = Parse::RecDescent->new($grammar);
$parser->match('sing aaaaaabbb song');
[download]

A lot of this code could be simplified and improved. I am neither a regex nor Parse::RecDescent guru. I did show how either could work.

Once you have a string of a's followed by one or more b's ($item[1]), you only needed to calculate the desired string and assign it to $return. An explicit assignment to $return is not necessary as you could just let the last expression be returned as with Perl's subroutines.

Cheers - L~R

Comment on Re: Parse::RecDescent for simple syntax-directed translation Select or Download Code

Replies are listed 'Best First'.
Re^2: Parse::RecDescent for simple syntax-directed translation by tomazos (Deacon) on Jun 18, 2006 at 18:17 UTC
Thanks for answering my question. The regex solution is cool. I should have qualified my statement by saying that standard regular grammars cannot handle this sort of pattern, whereas Perl's regexes can do everything and anything. In fact embedded actions within a Perl regex can do anything Perl can do - Therefore Perl regex's can do anything Perl can do. :) I guess I can see from your use of Parse::RecDescent, Update: And from re-reading the very long manual last night, that the answer to my question is something like: `my $grammar = q{ match : part(s) { print join('', @{$item[1]}) } part : AnB part : 'a' part : /[^a]+/ AnB : 'a' AnB 'b' { 'c' . $item[2] . 'd' } AnB : 'ab' { 'cd' } }` [download] -Andrew.	[reply] [d/l]
Re^3: Parse::RecDescent for simple syntax-directed translation by ikegami (Patriarch) on Jun 20, 2006 at 23:52 UTC
Close. Your grammar strips out all whitespace. Replace `match : part(s) { print join('', @{$item[1]}) }` with `match : <skip:''> part(s) { print join('', @{$item[2]}) }` A slight improvement is to replace `match : <skip:''> part(s) { print join('', @{$item[2]}) }` with `process : <skip:''> part(s) { join('', @{$item[2]}) }` so you can do `$filter = Parse::RecDescent->new($grammar);` `print $filter->process('eeeeaaaabbbeeee');`	[reply] [d/l] [select]