Why is this regular expression faster?

Question

I'm writing a Telnet client of sorts in C# and part of what I have to parse are ANSI/VT100 escape sequences, specifically, just those used for colour and formatting (detailed here).

One method I have is one to find all the codes and remove them, so I can render the text without any formatting if needed:

    
public static string StripStringFormating(string formattedString)
{
    if (rTest.IsMatch(formattedString))
        return rTest.Replace(formattedString, string.Empty);
    else
        return formattedString;
}

I'm new to regular expressions and I was suggested to use this:

static Regex rText = new Regex(@"\e\[[\d;]+m", RegexOptions.Compiled);

However, this failed if the escape code was incomplete due to an error on the server. So then this was suggested, but my friend warned it might be slower (this one also matches another condition (z) that I might come across later):

static Regex rTest = 
              new Regex(@"(\e(\[([\d;]*[mz]?))?)?", RegexOptions.Compiled);

This not only worked, but was in fact faster to and reduced the impact on my text rendering. Can someone explain to a regexp newbie, why? :)

Answer 1

Do you really want to do run the regexp twice? Without having checked (bad me) I would have thought that this would work well:

public static string StripStringFormating(string formattedString)
{    
    return rTest.Replace(formattedString, string.Empty);
}

If it does, you should see it run ~twice as fast...