用正则表达式匹配parens是很有可能的。

这是一个Perl脚本，将解析任意深度匹配的parens。

#!/usr/bin/perl
$re = qr  /
     (                      # start capture buffer 1
        \(                  #   match an opening paren
        (                   # capture buffer 2
        (?:                 #   match one of:
            (?>             #     don't backtrack over the inside of this group
                [^()]+    #       one or more 
            )               #     end non backtracking group
        |                   #     ... or ...
            (?1)            #     recurse to opening 1 and try it again
        )*                  #   0 or more times.
        )                   # end of buffer 2
        \)                  #   match a closing paren
     )                      # end capture buffer one
    /x;


sub strip {
    my ($str) = @_;
    while ($str=~/$re/g) {
        $match=$1; $striped=$2;
        print "$match\n";
        strip($striped) if $striped=~/\(/;
        return $striped;
    }
}

while(<DATA>) {
    print "start pattern: $_";
    while (/$re/g) { 
        strip($1) ;
    }
}   

__DATA__
"(apple + (-0.5)) * (boy - 1)"
"((((one)two)three)four)x(one(two(three(four))))"
"a) * (b + c) / (d"
"-a * (b / 1.50)"

输出：

start pattern: "(apple + (-0.5)) * (boy - 1)"
(apple + (-0.5))
(-0.5)
(boy - 1)
start pattern: "((((one)two)three)four)x(one(two(three(four))))"
((((one)two)three)four)
(((one)two)three)
((one)two)
(one)
(one(two(three(four))))
(two(three(four)))
(three(four))
(four)
start pattern: "a) * (b + c) / (d"
(b + c)
start pattern: "-a * (b / 1.50)"
(b / 1.50)