Domanda

I have issue with following example:

import java.util.regex.*;
class Regex2 {
    public static void main(String[] args) {
        Pattern p = Pattern.compile(args[0]);
        Matcher m = p.matcher(args[1]);
        boolean b = false;
        while(b = m.find()) {
            System.out.print(m.start() + m.group());
        }
    }
}

And the command line:

java Regex2 "\d*" ab34ef

Can someone explain to me, why the result is: 01234456

regex pattern is d* - it means number one or more but there are more positions that in args[1],

thanks

È stato utile?

Soluzione

\d* matches 0 or more digits. So, it will even match empty string before every character and after the last character. First before index 0, then before index 1, and so on.

So, for string ab34ef, it matches following groups:

Index    Group
  0        ""  (Before a)
  1        ""  (Before b)
  2        34  (Matches more than 0 digits this time)
  4        ""  (Before `e` at index 4)
  5        ""  (Before f)
  6        ""  (At the end, after f)

If you use \\d+, then you will get just a single group at 34.

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