提取每个级别的括号的内容

Question

I am converting a SMAPI grammar to JSGF . They are pretty similar grammars used in different speech recognition systems. SMAPI uses a question mark they way the rest of the world does, to mean 0 or 1 of the previous thing. JSGF uses square brackets for this. So, I need to convert a string like stuff? to [stuff] , and parenthesized strings like ((((stuff)? that)? I)? like)? to [[[[stuff] that] I] like] . I have to leave alone strings like ((((stuff) that) I) hate) . As Qtax pointed out, a more complicated example would be (foo ((bar)? (baz))?) being replaced by (foo [[bar] (baz)]) .

Because of this, I have to extract every level of a parenthesized expression, see if it ends in a question mark, and replace the parens and question mark with square braces if it does. I think Eric Strom's answer to this question is almost what I need. The problem is that when I use it, it returns the largest matched grouping, whereas I need to do operations on each individual groupings.

This is what I have so far: s/( \\( (?: [^()?]* | (?0) )* \\) ) \\?/[$1]/xg . When matched with ((((stuff)? that)? I)? like)? , however, it produces only [((((stuff)? that)? I)? like)] . Any ideas on how to do this?

I

Answer 1

You'll also want to look at ysth's solution to that question , and use a tool that is already available to solve this problem:

use Text::Balanced qw(extract_bracketed);
$text = '((((stuff)? that)? I)? like)?';

for ($i=0; $i<length($text); $i++) {
    ($match,$remainder) = extract_bracketed( substr($text,$i), '()' );
    if ($match && $remainder =~ /^\?/) {
        substr($text,$i) =
            '[' . substr($match,1,-1) . ']' . substr($remainder,1);
        $i=-1; # fixed
    }
}

Answer 2

In older Perl versions (pre 5.10), one could have used code assertions and dynamic regex for this:

 ...
 my $s = '((((stuff)? that)? I)? like)?';

 # recursive dynamic regex, we need
 # to pre-declare lexical variables
 my $rg;

 # use a dynamically generated regex (??{..})
 # and a code assertion (?{..})
 $rg = qr{
          (?:                       # start expression
           (?> [^)(]+)              # (a) we don't see any (..) => atomic!
            |                       # OR 
           (                        # (b) start capturing group for level
            \( (??{$rg}) \) \?      # oops, we found parentheses \(,\) w/sth 
           )                        # in between and the \? at the end
           (?{ print "[ $^N ]\n" }) # if we got here, print the captured text $^N
          )*                        # done, repeat expression if possible
         }xs;

 $s =~ /$rg/;
 ...

during the match, the code assertion prints all matches, which are:

 [ (stuff)? ]
 [ ((stuff)? that)? ]
 [ (((stuff)? that)? I)? ]
 [ ((((stuff)? that)? I)? like)? ]

To use this according to your requirements, you could change the code assertion slightly, put the capturing parentheses at the right place, and save the matches in an array:

 ...
 my @result;
 my $rg;
 $rg = qr{
          (?:                      
           (?> [^)(]+)             
            |                      
            \( ( (??{$rg}) ) \) \?  (?{ push @result, $^N })
          )*                     
         }xs;

 $s =~ /$rg/ && print map "[$_]\n", @result;
 ...

which says:

 [stuff]
 [(stuff)? that]
 [((stuff)? that)? I]
 [(((stuff)? that)? I)? like]

Regards

rbo

Answer 3

You could solve it in a couple of ways, simplest being just executing your expression till there are no more replacements made. Eg:

1 while s/( \( (?: [^()?]* | (?0) )* \) ) \?/[$1]/xg;

But that is highly inefficient (for deeply nested strings).

You could do it in one pass like this instead:

s{
  (?(DEFINE)
    (?<r>   \( (?: [^()]++ | (?&r) )*+ \)   )
  )

  ( \( )
  (?=   (?: [^()]++ | (?&r) )*+ \) \?   )

  |

  \) \?
}{
  $2? '[': ']'
}gex;