I am writing a parser that takes a user-specified string and evaluates the tokens within it. A token in this case meaning an expression grouped within parentheses. The idea is to take one string, and recursively replace each token with its evaluated value. So, (2 > 1) would be replaced with 1. This would allow any combination of tokens to be grouped into a string of nested parentheses, and then be parsed. The code that follows does this quite well...up until the very last replacement, when there is just one token left. And as a result, I'm rather puzzled. This is a test case of code, written to output what it's doing. 
#! /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 $val = (eval $1) ? 1 : 0;
    print "Case to be evaluated = $1\n";
    print "Evaluated value = $val\n";
    $string =~ s/\($1\)/$val/;
    print "Parsed string = $string\n";
}
print "Result: $string\n";

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The code as shown above will end with 1|| 0), instead of just 1. However, if I change the | to & in the test string? It works fine. Which puzzles me to no end. So Great Monks, whose wisdom far outstrips mine own...how do I make this work more consistently? ~Moe~

In reply to Parsing parenthetical arguments recursively by Moe

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