Delimiter in Scanner Java confusion

Question

According to Java API Scanner uses delimiters to break the whole input into tokens. I am trying to understand the tokens and delimiters. I was doing this program and hit a confusion

import java.util.Scanner;

public class Test {
    public static void main(String[] args) {
        Scanner s = null;
        try {
            s = new Scanner(System.in);
            s.useDelimiter("A");
            System.out.println("1 " + s.next().length());
            System.out.println("2 " + s.next().length());
            System.out.println("3 " + s.next().length());
            System.out.println("4 " + s.next().length());
        } finally {
            if (s != null) {
                s.close();
            }
        }
    }
}

When I use the input AAAAAasdf I get the following output.

1 0
2 0
3 0
4 0

I can understand this output as the length of tokens is zero between the delimiters hence all are zero but when I use the default delimiters and give the input as

_____aaa\n ->Replace underscore by space and \n by me hitting enter in eclipse console.

For this I am getting the output as

1 3

which I cannot understand. I have given 5 spaces so there should be 4 tokens of 0 lengths between them. Why not? What am I missing here?

Answer 1

useDelimiter takes a regular expression pattern. The default pattern is

private static Pattern WHITESPACE_PATTERN = Pattern.compile(
                                            "\\p{javaWhitespace}+");

Which will match any amount of contiguous whitespace. If you want the delimiter to match any amount of contiguous A try something like

s.useDelimiter("[A]+");

Read these: http://docs.oracle.com/javase/7/docs/api/java/util/Scanner.html#useDelimiter(java.lang.String) http://docs.oracle.com/javase/7/docs/api/java/util/Scanner.html#reset()