matcher.matches()

[Garrett Smith]

Can someone explain this?

    Pattern p = Pattern.compile("\\d+(x)?");
    Matcher m = p.matcher("1x");
   // m.matches();
    System.out.println(m.group(0));

[Garrett-Smiths-Computer:~/Documents/java/patterns] garrett% java Patterns
Exception in thread "main" java.lang.IllegalStateException: No match found
at java.util.regex.Matcher.group(Matcher.java:353)
at Patterns.main(Patterns.java:54)

    Pattern p = Pattern.compile("\\d+(x)?");
    Matcher m = p.matcher("1x");
    m.matches();
    System.out.println(m.group(0));

[Garrett-Smiths-Computer:~/Documents/java/patterns] garrett% java Patterns
1x
Unless I call m.matches, the pattern fails. Why? Is this a bug?
[ January 18, 2003: Message edited by: Garrett Smith ]

[Jim Yingst]

No, this is normal behavior. A Matcher does not automatically try to match a given pattern until you call either matches() or find() - because it does not know in advance whether you want it to match the entire input at once (as for matches()) or just look for a match somewhere within the input (as for find()). Thus, you must call either matches() or find() (and check that the return value is true) before trying to use group(int).

[Garrett Smith]

[B]group[/B]
public String group()
    Returns the input subsequence matched by the previous match.

Since no previous match exists, then the IllegalStateException is thrown.
matches() tries to match a pattern, thereby changing the state of the matcher.