Is regex in Java anchored by default with both a ^ and $ character?

Question

From my understanding of regular expressions, the string "00###" has to match with "[0-9]", but not with "^[0-9]$". But it doesn't work with Java regexp's.

After some investigating of this problem I found the following information (http://www.wellho.net/solutions/java-regular-expressions-in-java.html):

It might appear that Java regular expressions are default anchored with both a ^ and $ character.

Can we be sure that this is true for all versions of JDK? And can this mode be turned off (i.e. to disable default anchoring with ^ and $)?

Answer 1

As the article you linked to explains, it depends on the function you call. If you want to add ^ and $ by default, use String#matches or Matcher#matches. If you don't want that, use the Matcher#find method instead.

import java.util.regex.*;  public class Example {     public static void main(String[] args)     {         System.out.println("Matches: " + "abc".matches("a+"));          Matcher matcher = Pattern.compile("a+").matcher("abc");         System.out.println("Find: " + matcher.find());     } } 

Output:

Matches: false Find: true 

Answer 2

Yes, matches() always acts as if the regex were anchored at both ends. To get the traditional behavior, which is to match any substring of the target, you have to use find() (as others have already pointed out). Very few regex tools offer anything equivalent to Java's matches() methods, so your confusion is justified. The only other one I can think of offhand is the XML Schema flavor.