RFC: Perl regex to validate arithmetic expressions

[Serge314]

Sorry, I'm not a native English speaker. Professional Perl programmers doubt that with one regex can validate arithmetic expressions, because of unlimited level of nested parentheses, and they use special Perl modules for this purpose. I wrote such regex. To do this, I used simple and elegant method by which you can write a similar complex expressions. I wrote an article about it, but unfortunately not in English ...
Here's my example:

#!perl -w
use strict;
use re 'eval';

# Validation of arithmetic expressions like +1*(2-3/(4*5))
# Author: Serge Melnikov (www.cronc.com)
 
my @testexpr=(
# Valid arithmetic expressions:
'+3',
'-(3/4+(-2-3)/3)',
'(-((3)))',
'-(+1+2)*(3/(1-2))/((-3))',
# Invalid (from the point of view of a human or mathematical notation)
+ arithmetic expressions:
'2*-3',           # in math we must write 2*(-3)
'-(3/4+(-2-3)/3',
'(-((3)+))');
 
my $number=qr/\d+/; # the regex for a term (number/variable/function c
+all)
my $arithm;
# The x modifier used only to split the long regex to publish it
$arithm=qr#[+-]?(?:$number|\((??{$arithm})\))(?:[*/](?:$number|\((??{$
+arithm})\)))*
   (?:[+-](?:$number|\((??{$arithm})\))(?:[*/](?:$number|\((??{$arithm
+})\)))*)*#x;
 
print /^$arithm$/ ? "Valid: $_\n" : "Invalid: $_\n" for (@testexpr);

[JavaFan]

my $pat = qr {
    (?(DEFINE)
       (?<number>    (?: [0-9]+ ))
       (?<sign>      (?: [-+] ))
       (?<expr>      (?: (?&term) (?: \s* [-+] \s* (?&expr))? ))
       (?<term>      (?: (?&factor) (?: \s* [/*] \s* (?&term))? ))
       (?<factor>    (?: (?&number) |
                         (?&sign) \s* (?&factor) |
                        \( \s* (?&expr) \s* \)))
    )
    (?&expr)
}x;

That will even parse 2*-3, and doesn't require (??{ }), or use re 'eval';.

[Fox]

That's pretty cool, I didn't knew about the re 'eval' and this just opened my mind to a whole new class of solutions using these 'recursive' regexes.

I just don't like the lack of whitespace, maybe you could allow it by adding a s/\s//g before the main test ? not sure if it will break the test.

[mellon85]

You actually can parse an arithmetic expression with a "regular" expression if it uses a stack automata. Just think that a regular language can't even check if there is a correct number of parenthesis if the maximum depth is undefined, as such, strictly speaking that is not so much regular...

[JavaFan]

Since 5.10, Perl regular expressions can match anything that's matched by a context-free grammar - and then some (backreferences add even more power). Matching balanced parenthesis is easy in 5.10 and later: /^(\((?1)*\))$/.

[spx2]

there's also the option of using Parse::Yapp and validating arithmetic expressions. In general when "discovering" Perl regexes you think they are so powerful that you can use them for a lot of stuff, but if all you have is a hammer, everything looks like a nail.