Regular expression to apply backspace characters

Question

I have a string coming from a telnet client. This string contains backspace characters which I need to apply. Each backspace should remove one previously typed character.

I'm trying to do this in a single replace using regular expression:

string txt = "Hello7\b World123\b\b\b";
txt = Regex.Replace(txt, ".\\\b", "", RegexOptions.ECMAScript);

Which results in "Hello World12". Of course, I want "12" to be removed too, but it obviously doesn't match my expression.

In some way, it should repeat replacing until there are no more matches. Any ideas on how to achieve this with a single regular expression?

Answer 1

This is basically a variant of How can we match a^n b^n with Java regex?, so we could reuse its answer there:

var regex = new Regex(@"(?:[^\b](?=[^\b]*((?>\1?)[\b])))+\1");
Console.WriteLine(regex.Replace("Hello7\b World123\b\b\b", ""));

Additionally, the .NET regex engine supports balancing groups, so we could use a different pattern:

var regex = new Regex(@"(?<L>[^\b])+(?<R-L>[\b])+(?(L)(?!))");

(This means:

1. Match one or more non-backspaces, assigning them with the name "L",
2. then followed one or more backspaces, assigning them with the name "R", with the condition that every "R" must have one corresponding "L",
3. if there are any "L"s left, abandon the match (as (?!) matches nothing).

)