Regex Question - Quantifiers' behavior.

[Chan Ag]

My Regex code is given below. Could somebody please help me understand the output ( specifically the highlighted parts of the output). If my string to be matched is only 3 characters long, how can I have a blank string starting at position 3? Would I see this behavior only when the quantifier is a '0 or' more quantifier? Also when I have a pattern that constitutes a zero or more quantifier plus the reluctant behavior, I only get blank strings while my original string is "abc" ( not three blanks or four?). Thanks for your help.

package RegexExamples;
import java.util.regex.Pattern;
import java.util.regex.Matcher;


public class RegexExample1 {
    
    public static void main(String[] args) {
        
        Pattern[] pattern = new Pattern[7];
        pattern[0] = Pattern.compile("[\\d]");
        pattern[1] = Pattern.compile("[\\d]+");
        pattern[2] = Pattern.compile("[\\d]+?");
        pattern[3] = Pattern.compile("[\\d]*");
        pattern[4] = Pattern.compile("[\\d]*?");
        pattern[5] = Pattern.compile("[\\d]?");
        pattern[6] = Pattern.compile("[\\d]??");
        Matcher[] matcher = new Matcher[7];
        for (int i =0; i<7; i++)
        {
                matcher[i] = pattern[i].matcher("123");
                System.out.println("\n ------------------------------------------\n");
                System.out.println("Pattern is " + matcher[i].pattern());
        
                while (matcher[i].find()) {
                    System.out.println(matcher[i].start() + " String = " + matcher[i].group());
        }
        
    }
}
}

My output listing is as follows.

run:

------------------------------------------

Pattern is [\d]
0 String = 1
1 String = 2
2 String = 3

------------------------------------------

Pattern is [\d]+
0 String = 123

------------------------------------------

Pattern is [\d]+?
0 String = 1
1 String = 2
2 String = 3

------------------------------------------

Pattern is [\d]*
0 String = 123
3 String =

------------------------------------------

Pattern is [\d]*?
0 String =
1 String =
2 String =
3 String =

------------------------------------------

Pattern is [\d]?
0 String = 1
1 String = 2
2 String = 3
3 String =

------------------------------------------

Pattern is [\d]??
0 String =
1 String =
2 String =
3 String =
BUILD SUCCESSFUL (total time: 0 seconds)

[Tony Docherty]

Pattern is [\d]*
0 String = 123
3 String =

* matches 0 or more times in a greedy fashion, so it matches all the digits and no digits at the end of the string.

Pattern is [\d]*?
0 String =
1 String =
2 String =
3 String =

*? matches 0 or more times in a lazy fashion, so it matches before each digit and no digits at the end of the string.

Pattern is [\d]?
0 String = 1
1 String = 2
2 String = 3
3 String =

? matches 0 or one time in a greedy fashion, so it matches each of the digits and no digits at the end of the string.

Pattern is [\d]??
0 String =
1 String =
2 String =
3 String =

*? matches 0 or one time in a lazy fashion, so it matches before each digit and no digits at the end of the string.

[Chan Ag]

Thanks, Tony.

I didn't know it was an empty string and that in the lazy behavior the match is done before the digit.