can some one explain me the output of this program?

[Pooja Oza]

   

Given:
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
What is the result?
A. 234
B. 334
C. 2334
D. 0123456
E. 01234456
F. 12334567
G. Compilation fails
Answer:
® 3 E is correct. The \d is looking for digits. The * is a quantifier that looks for 0 to many
occurrences of the pattern that precedes it. Because we specified *, the group() method
returns empty Strings until consecutive digits are found, so the only time group() returns
a value is when it returns 34 when the matcher finds digits starting in position 2. The
start() method returns the starting position of the previous match because, again,
we said find 0 to many occurrences.
 

[Ulf Dittmer]

That's a FAQ: http://faq.javaranch.com/java/ScjpFaq#kb-regexp

In the future, please QuoteYourSources when posting questions like these.

[Pooja Oza]

Thanks Ulf for your help