Regex Question with * quantifier from K&B

[Raj Purushotham]

Hi,

I am not clear on the Regex question (page 508 Question 1) from K&B 5 book

To uderstad I tried writing a similar program.

import java.util.regex.*;
class Regex{
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.println(m.start()+ "space" + m.group());

}
	}// end of main
}
 

Lets say I give the followig commad line Java Regex "\d*" a
I get

0space
1space

I see that it is searchig for digits ad it finds 0 matches first time and prints postion 0 (search index 0). Then because we said * the start method returns to the previous index 0 now how come it goes to the postion 1 when there is only 1 charecter in the search strig.

[Abimaran Kugathasan]

UseCodeTags when you post a code. Please Edit it and put code tags. Welcome to JavaRanch!

[Abimaran Kugathasan]

OK, What would happen you put + instead of *?
* - zero or more times
+ - one or more times.

Got it?

[Raj Purushotham]

Hi Abimaran,

Thank you for the respose. I tried with + istead of * and the results are same.

I can see that if I use "\d?" for the source a23a12 it gives me what I suspect

1space23
4space12

How ever if I use "\d*" for the source a23a12 it gives me the following:
0space
1space23
3space
4space12
6space

"6space" is not clear to me.