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

by Limbic~Region (Chancellor)
on Jun 18, 2006 at 15:16 UTC ( [id://556097] : note . print w/replies, xml ) Need Help??


in reply to 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;

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');
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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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' } }

    -Andrew.

      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');