in reply to Parsing parenthetical arguments recursively

And a colleague looking over my shoulder points and says: You have to escape your |'s. And I stare. And I splutter. And I try what he suggests, and it appears to work miraculously well.

If the Great Holy Monks have further input, I am forever grateful.

Revised code: 
#! /usr/bin/perl
#sample string.
$string = "((2 > 1) | ('word' eq 'toy'))";
#next line converts user-specified operators into Perl
#logical operators. 
$string =~ s/([&|]{1})/$1$1/g;
print "Initial parsed string = $string\n";
#Look ahead assertion.
while ($string =~ /\(((?:[^()])*?)\)/) {
    my $assert = $1;
    my $val = (eval $1) ? 1 : 0;
    print "Case to be evaluated = $1\n";
    print "Evaluated value = $val\n";
    $assert =~ s/\|\|/\\\|\\\|/g;
    $string =~ s/\($assert\)/$val/;
    print "Parsed string = $string\n";
}
print "Result: $string\n";

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Re: Re: Parsing parenthetical arguments recursively
by Anonymous Monk on Aug 15, 2003 at 19:38 UTC
    Its possible you want to quotemeta() $assert instead, but I could see it going either way.
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Re: Re: Parsing parenthetical arguments recursively
by waswas-fng (Curate) on Aug 15, 2003 at 22:29 UTC
    So you don't want to test and reject unbalanced parens? for instance: 
    
$string = "((2 > 1) | ('word' eq 'toy')))";

    [download]
    If you did you may want to use one of the balanced modules to verify balance of the parens.

    -Waswas
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[d/l]
      The same code(now working) is being used to accomplish that, actually. I use the same regexp when the user supplies the expression to verify its validity, without doing any evaluation on it:

      while ($tmp =~ /\(((?:[^()])*?)\)/) {
        my $assert = $1;
        $assert =~ s/\|/\\\|/g;
        $tmp =~ s/\($assert\)/$1/;
}
($tmp =~ /\(|\)/) ? print "Don't match.\n" : print "Match.\n";

      [download]

      Just as a purely hypothetical example. Though, I will definitely look into those modules, as I'd not heard of them before. Thank you for the link.

      ~Moe~
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