http://qs1969.pair.com?node_id=555986

tomazos has asked for the wisdom of the Perl Monks concerning the following question:

I'm trying to work out how to use Parse::RecDescent and am a little overwhelmed.

I can see how to use it as a recognizer, but to do a simple syntax-directed translation, all of the *items, $return, $text, real return value, etc, stuff has thrown me off.

Let's say you want a translator for languages over ('a','b','c','d','e') such that any balanced pairs of equal length sequences of a's and b's are replaced by c's and d's. 

my $start = "eeeeaaaabbbeeee";
my $end = translate($start);
# $end eq "eeeeacccdddeeee"  (aaabbb -> cccddd)

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(Side note: This is the "classic example" of something that can't be done with Update: regular grammars, 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).)

The syntax-directed translation (in pseudo-code) would be: 

start -> part(s)    { start.t := join ('', part(s).t) }

part -> AnB         { part.t := AnB.t }
part -> 'a'         { part.t := 'a' }
part -> /[^a]+/     { part.t := /[^a]+/.t }

AnB -> 'a' AnB 'b'  { AnB.t := 'c' . AnB.t . 'd' }
AnB -> 'ab'         { AnB.t := 'cd' }

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Any ideas on how this translates into Parse::RecDescent? Is there a more appropriate parsing module to use for cases where the input language is very similiar to the output language?

-Andrew.

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Re: Parse::RecDescent for simple syntax-directed translation
by Limbic~Region (Chancellor) on Jun 18, 2006 at 15:16 UTC
    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');

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    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

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      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.

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[d/l]

        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');

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