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String split() method bug?

Posted on 2004-04-05
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3,045 Views
Last Modified: 2007-12-19
Consider the following:

String s = "#1#sdsd###2#dsfdf##";
System.out.println(s);
String sa [] = s.split("#");
for (int i=0; i<sa.length; i++)
{
      System.out.println("token[" + i + "]: " + sa[i]);
}

It only returns 7 tokens instead of (the expected?) 8 tokens. Is this a bug or am I missing something?  I looked at java bug database but couldn't find anything.

Any thoughts?

  Thank you.
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Question by:girionis
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9 Comments
 
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Expert Comment

by:Webstorm
ID: 10757212
From API javadoc:
  " Trailing empty strings are therefore not included in the resulting array. "


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Webstorm earned 66 total points
ID: 10757215
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Expert Comment

by:Webstorm
ID: 10757231
If you want the trailing empty strings :

String s = "#1#sdsd###2#dsfdf##";
System.out.println(s);
s=s+"end";
String sa [] = s.split("#");
for (int i=0; i<sa.length-1; i++)
{
     System.out.println("token[" + i + "]: " + sa[i]);
}
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Author Comment

by:girionis
ID: 10757246
Maybe next time I should read the javadocs first.. thank you :)
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Expert Comment

by:CEHJ
ID: 10757258
>> " Trailing empty strings are therefore not included in the resulting array. "

Do you want it to be?
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Expert Comment

by:Andy_Neodynium
ID: 10757264
Trailing empty strings are not returned. Look up the docs for java.lang.String.split(). Those last two '#'s won't return anything

"This method works as if by invoking the two-argument split method with the given expression and a limit argument of zero. Trailing empty strings are therefore not included in the resulting array. "
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Expert Comment

by:CEHJ
ID: 10757287
If you DO want it to be, here's the correct RE

String sa[] = s.split("(?=#)");
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Expert Comment

by:Webstorm
ID: 10757320
:-)
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Author Comment

by:girionis
ID: 10757490
Thank you everyone for your replies.

>If you DO want it to be, here's the correct RE
>
>String sa[] = s.split("(?=#)");

No I don't, the real problem was an IndexOutOfBoundsException and I just solved by iterating at the length - 1 of the array.
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