Regex return to pattern - Solved

[Rodrigo Tassini]

This is a example for SCJP for Java Platform, SE6 STUDY GUIDE by Richard F. Raposa and can be found at page 317.

 
public class RegexTest3 {

	public static void main(String[] args) {
		String regex = "[qrstuv]*.*ing";
		Pattern pattern = Pattern.compile(regex);
		
		String [] tests = {"ing", "ring", "trying" , "running",
				"beings"};
		
		for (String s: tests) {
			Matcher m = pattern.matcher(s);
			if (m.matches()) {
				System.out.println(s + " matches " + regex );
			}
		}

	}

}

In the books is write
The pattern matches any word that starts with any character between ‘ q ’ and ‘ v ’ (the [qrstuv] ) repeated any number of times (the “ * ” ), followed by any single character (the . ), and ending with the literal string “ ing ” . Using the tests array from the previous examples, the matches are
ring matches [qrstuv]*.ing
trying matches [qrstuv]*.ing


In my compilatin, the return was:
ing matches [qrstuv]*.*ing
ring matches [qrstuv]*.*ing
trying matches [qrstuv]*.*ing
running matches [qrstuv]*.*ing

Someone understand what's happened?

Thks in advance

[Henry Wong]

"*" does mean "repeated any number of times" including zero times. So, for the other two cases, where the word doesn't start with "q" to "v", the regex engine matches it zero times. This allows the "." to match the first character instead.

Henry

[Ankit Garg]

4 values match the pattern give, so if you are getting 4 matches as output, then its correct, the answer might be wrong in the book...

[Rodrigo Tassini]

Sorry guys, I'd make a mistake.

The correct pattern is

    
"[qrstuv]*.ing"

[Henry Wong]

There is no need to change the subject to solved... there is a "resolve" button below, that you can use to declare the topic as solved.

Henry