Question about String split() ... (was DOUBT)

[sravanthi pulukuri]

string str="aaaaaaaaabb";
String[]s=str.split(a{3});
system.out.println(s.length);
Can you please help me ou why the output is 4


[HENRY: Added better subject title]
[ May 07, 2007: Message edited by: Henry Wong ]

[Giovanni Murguia]

Hello!

I think the code is:

String str="aaaaaaaaabb";
String[]s=str.split("a{3}");
System.out.println(s.length);

You're using as separator 3 a's (aaa) so when it begins the search it uses the first 3 a's, the string before this would be an empty string. It removes those 3 a's and now search using "aaaaaabb". It does that with the other aaa's until the last string is aaab. Using split gives 2 tokens: an empty string and "bb". Try compiling this:

String str="aaaaaaaaabb";
String[]s=str.split("a{3}");
for(int i=0; i<s.length; i++) {
String tmp = s[i];
System.out.println(i+1+": "+tmp);
}
System.out.println("Total length: "+s.length);

Hope it helps.

Giovanni

[Jesse Custer]

The code you provided doesn't compile, please check your syntax before posting. This is the correct code with an added line so it's better to see what happens here. The regex you use is the same as "aaa".

public class Test{
public static void main(String args[]){
String str = "aaaaaaaaabb";
String[]s = str.split("a{3}");
System.out.println(s.length);
 
for(String temp:s) System.out.println(">" + temp + "<");
}
}

[Chandra Bhatt]

Does following code give any helping idea:

import java.util.regex.*;
 
public class DemoSplit1 {
 
public static void main(String... args) {
String str="aaaaaaaaaaaaaaaaaabb";
String[]s=str.split("a{3}");
 
for(String s1: s) {
System.out.println(">"+s1+"<");
}
System.out.println(s.length);

System.out.println("-------------------------------");
Pattern p1 = Pattern.compile("a{3}");
Matcher m1 = p1.matcher(str);

while(m1.find()) {
System.out.println(m1.start() + " >" + m1.group() + "<");
}

}
}
 

Output:
><
><
><
><
><
><
>bb<
7
-------------------------------
0 >aaa<
3 >aaa<
6 >aaa<
9 >aaa<
12 >aaa<
15 >aaa<



Thanks,

[John Stone]

Chandra:
try to change string in your example to:

String str="bbaaaaaaaaaaaaaaaaaa";

Results of split and Matcher are quite different.

[Chandra Bhatt]

Thanks John,

I see the difference; But I also tried this:

String str = "bbaaaaaaaaaaaaaaaaaab";

Output:
>bb<
><
><
><
><
><
>b<
7

Could you put some focus on that?


Thanks,
[ May 07, 2007: Message edited by: Chandra Bhatt ]

[sravanthi pulukuri]

Hi
chandra could you explain on the output you got
I guess one more has to come before <b>

[John Stone]

it depends on whether limit parameter was supplied.

from Pattern::split(...)

  if (limit == 0)
            while (resultSize > 0 && matchList.get(resultSize-1).equals(""))
                resultSize--; 

if there is no limit parameter, split will chop off all empty matches from the end, until non-empty one is found

Try the same string, but put extra parameter for limit (let's say 100) and you should get all matches.

[Chandra Bhatt]

I agree John,

public class TestSplit {
 
public static void main(String... args) {
 String str = "bbaaaaaaaaaaaaaaaaaaaaa"; ;
 [B]String[] sarr = str.split("a{3}",4);[/B] //Line #1

 for(String s:sarr) {
 System.out.println(">" + s + "<");
 }
}
}

Output:
0>bb<
1><
2><
3>aaaaaaaaaaaa<


Change line #1 with
String[] sarr = str.split("a{3}",10);

Output:
0>bb<
1><
2><
3><
4><
5><
6><
7><

and without passing the LIMIT the out:

0>bb<

It means without limit all the empty matches are chopped off.


Thanks,

[sravanthi pulukuri]

No chandra
Am not able to follow,please can you explian me from start
Thanks

[John Stone]

Originally posted by Chandra Bhatt:

It means without limit all the empty matches are chopped off.

.. from the end only.
"bbaaaaaaaaaaaaaaaaaabb" will have empty mathces and you don't need to set any limit.