How does MatchEvaluator in Regex.Replace work?

Question

This is the input string 23x * y34x2 . I want to insert " * " (star surrounded by whitespaces) after every number followed by letter, and after every letter followed by number. So my output string would look like this: 23 * x * y * 34 * x * 2 .

This is the regex that does the job: @"\\d(?=[az])|[az](?=\\d)" . This is the function that I wrote that inserts the " * " .

Regex reg = new Regex(@"\d(?=[a-z])|[a-z](?=\d)");   
MatchCollection matchC;
matchC = reg.Matches(input);
int ii = 1;
foreach (Match element in matchC)//foreach match I will find the index of that match
{
    input = input.Insert(element.Index + ii, " * ");//since I' am inserting " * " ( 3 characters )
    ii += 3;                                        //I must increment index by 3
}
return input; //return modified input

My question how to do same job using .net MatchEvaluator ? I am new to regex and don't understand good replacing with MatchEvaluator . This is the code that I tried to wrote:

{
    Regex reg = new Regex(@"\d(?=[a-z])|[a-z](?=\d)");
    MatchEvaluator matchEval = new MatchEvaluator(ReplaceStar);
    input = reg.Replace(input, matchEval);
    return input;
}
public string ReplaceStar( Match match )
{
    //return What??
}

Answer 1

A MatchEvaluator is a delegate that takes a Match object and returns a string that should be replaced instead of the match. You can also refer to groups from the match. You can rewrite your code as follows:

string input = "23x * y34x2";
Regex reg = new Regex(@"\d(?=[a-z])|[a-z](?=\d)");
string result = reg.Replace(input, delegate(Match m) {
    return m.Value + " * ";
});

To give an example of how this works, the first time the delegate is called, Match parameter will be a match on the string "3" . The delegate in this case is defined to return the match itself as a string concatenated with " * " . So the first "3" is replaced with "3 * " .

The process continues in this way, with delegate being called once for each match in the original string.