Zero-length matches in Java Regex

Question

My code :

Pattern pattern = Pattern.compile(\"a?\");
Matcher matcher = pattern.matcher(\"ababa\");
while(matcher.find()){
   System.out.println(matcher.star         


        

Answer 1

The ? is a greedy quantifier, therefore it will first try to match the 1-occurence before trying the 0-occurence. In you string,

1. it starts with the first char 'a' and tries to match agains the 1-occurence. The 'a' char matches and so it returns the first result you see
2. then it moves forward and find a 'b'. The 'b' char does not match your regexp 1-occurence, so the engine backtracks and attempt to match a 0-occurence. Result is that the empty string is matched--> you get your second result.
3. then it moves ahead of b since no more matches are possible there and it starts again with your second 'a' char.
4. etc... you get the point...

It is a bit more complicated than that but that is the main idea. When the 1-occurence cannot match, it will then try with the 0-occurence.

As for the values of start, end and group, they will be where the match starts, ends and the group is what has been matched, so in the first 0-occurence match of your string, you get 1, 1 and the emtpy string. I am not sure this really answers your question.