backreference

I need to match a regex that uses backreferences (e.g. \1) in my Go code. That's not so easy because in Go, the official regexp package uses the RE2 engine , one that have chosen to not support backreferences (and some other lesser-known features) so that there can be a guarantee of linear-time execution, therefore avoiding regex denial-of-service attacks . Enabling backreferences support is not an option with RE2. In my code, there is no risk of malicious exploitation by attackers, and I need backreferences. What should I do? Regular Expressions are great for working with regular grammars,

Can you use backreferences in a lookbehind? Let's say I want to split wherever behind me a character is repeated twice. String REGEX1 = "(?<=(.)\\1)"; // DOESN'T WORK! String REGEX2 = "(?<=(?=(.)\\1)..)"; // WORKS! System.out.println(java.util.Arrays.toString( "Bazooka killed the poor aardvark (yummy!)" .split(REGEX2) )); // prints "[Bazoo, ka kill, ed the poo, r aa, rdvark (yumm, y!)]" Using REGEX2 (where the backreference is in a lookahead nested inside a lookbehind) works, but REGEX1 gives this error at run-time: Look-behind group does not have an obvious maximum length near index 8 (?<=(.)

You can backreference like this in JavaScript: var str = "123 $test 123"; str = str.replace(/(\$)([a-z]+)/gi, "$2"); This would (quite silly) replace "$test" with "test". But imagine I'd like to pass the resulting string of $2 into a function, which returns another value. I tried doing this, but instead of getting the string "test", I get "$2". Is there a way to achieve this? // Instead of getting "$2" passed into somefunc, I want "test" // (i.e. the result of the regex) str = str.replace(/(\$)([a-z]+)/gi, somefunc("$2")); Like this: str.replace(regex, function(match, $1, $2, offset, original)

Situation I want to use preg_replace() to add a digit '8' after each of [aeiou] . Example from abcdefghij to a8bcde8fghi8j Question How should I write the replacement string? // input string $in = 'abcdefghij'; // this obviously won't work ----------↓ $out = preg_replace( '/([aeiou])/', '\18', $in); This is just an example , so suggesting str_replace() is not a valid answer. I want to know how to have number after backreference in the replacement string. The solution is to wrap the backreference in ${} . $out = preg_replace( '/([aeiou])/', '${1}8', $in); which will output a8bcde8fghi8j See the