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quantifier

Preethi Dev
Ranch Hand

Joined: Sep 07, 2008
Posts: 265
import java.io.*;
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.print(m.start() + m.group());
}
}
}

command line: java regex "\d*" ab34ef
output: 01234456

i got this from K&B 5 study guide(self test question).

i am not getting how the '*'(quantifier) and 'm.group()' works here?
please anyone explain me...


Preparing Scjp5
Bob Ruth
Ranch Hand

Joined: Jun 04, 2007
Posts: 320
That one caused me trouble too.

The trick is in what "0 or more" means and what it does.

string "AB34EF"
pattern "\d*"

Okay, '*' says search for "0" or more numeric digits. As it steps through the string, checking the digits it will return with a match for every check that returns 0 or more digits.

When you check 'A' there are 0 digits so matcher returns 0 for the index and the group() is null.

When you check 'B' there are 0 digits so matcher returns 1 for the index and the group() is null.

When you check '3' there is a digit but we need to keep searching so no return is made.

When you check '4' there is a digit and we need to continue searching...

When you check the 'E' that ends the sequence of digits so two things happen
For the digits we found, matcher returns 2 for the index and the group() is "34". Because the match used indexes 2 and 3, the next search will start at index 4 and it has 0 digits so matcher returns 4 and group() as null.

When you check 'F' 0 digits are found so matcher returns with 5 for the index and group() is null.

One more check is made to find the end of the string and it returns 6 as the index and I imagine null as the string.

As an intermediate step to illustrate how the answer is formed let's stack all of these together as follows...

0null1null2344null5null6null

Since null is actuall "no string at all" let's remove all of the nulls:

01234456

I may be a little fuzzy on what happens at the end of the string (the index 6) but I believe this reasonably well explains what happens on this question. I hope that it will be of some help.

Bob

(edited to correct typos)
[ October 06, 2008: Message edited by: Bob Ruth ]

------------------------
Bob
SCJP - 86% - June 11, 2009
Thomas Thevis
Ranch Hand

Joined: Sep 02, 2008
Posts: 87
Originally posted by Bob Ruth:
When you check 'A' there are 0 digits so matcher returns 0 for the index and the group() is null.

Well, not exactly. 'group()' returns the string that was matched by the previous 'find()'. In this case, it is the empty string ("") which is really different from 'null'.

Regards,
Thomas


SCJP 5.0, SCJD in progress
Sandhya Bhaskara
Greenhorn

Joined: Jun 29, 2005
Posts: 23
I understood '0123445' part. I did not get how there can be a 6 when the last index of the string passed is 5. Totally confused about how start() works when it reaches end of the string


Sandhya Bhaskara<br />SCJP 1.4,SCWCD 1.4,SCBCD 1.3,SCJP 6
Sandhya Bhaskara
Greenhorn

Joined: Jun 29, 2005
Posts: 23
Probably m.find()it returns a true even after the end of string. Still confused..
Bob Ruth
Ranch Hand

Joined: Jun 04, 2007
Posts: 320
Thomas, looking back I agree. Thanks for the correction. (after all I think if group returned null it would probably PRINT null!)
Preethi Dev
Ranch Hand

Joined: Sep 07, 2008
Posts: 265
Hi Bob

i got some what cleared,but how that 6 comes at last?is it taking the next index also for matching?
marc weber
Sheriff

Joined: Aug 31, 2004
Posts: 11343

This question is addressed in our SCJP FAQ.


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vipin jain
Ranch Hand

Joined: Aug 24, 2008
Posts: 122
Hi Bob

i got some what cleared,but how that 6 comes at last?


Best Regards,<br />Vipin<br />MCA, SCJP5, SCWCD in progress
Ulf Dittmer
Marshal

Joined: Mar 22, 2005
Posts: 41186
    
  45
but how that 6 comes at last?

The FAQ entry marc linked to explains that.


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