Hi Everyone, Following is one question from K&B SCJP 5 book. I don't understand how this e answer is correct. 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.
Once at index 2, group brings 34. How come the rest of the result is arrived. Only until 5 are the index available, and if it prints empty strings again, how come it gets to 56 in the end? Thanks and regards, Subha. [ February 27, 2006: Message edited by: subathra sangameswaran ]
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