Efficiency with java strings

Question

When we extend DefaultHandler, we usually override the characters function.

I was wondering which would be a more efficient way to extract the String supplied...

Would it be using a for loop and a StringBuilder?

@Override
public void characters(char[] ch, int start, int length) throws SAXException {
    StringBuilder sb = new StringBuilder(length); 
    for(int i=start; i<start+length; i++)
        sb.append(ch[i]);
    String values = sb.toString();
}

Would it be using a simple substring?

@Override
public void characters(char[] ch, int start, int length) throws SAXException {
    String values = (new String(ch)).substring(start, start + length);
}

Is there another approach?

Note that the #characters(char[] ch, int, int) method may be called several times in a row and you should append character data in a StringBuilder class member and not process it until you hit the relevant #endElement(String,String,String) call. Appending chunks of characters should be done by calling StringBuilder#append(char[], int, int). — Jens
– Jens, Commented May 28, 2012 at 11:17
@Jens You are correct! I have been working a lot with feeds and I noticed this behavior. The problem is that it will be too messy to use a StringBuilder to append (especially for the part of maintaining reference). I prefer to use string concatenation in this case. — Sherif elKhatib
– Sherif elKhatib, Commented May 28, 2012 at 13:15
If you are implementing ContentHandlers manually you should probably consider using some sort of framework - even a simple one like the built in android.sax would probably simply things for you. — Jens
– Jens, Commented May 28, 2012 at 13:19
I have worked on a small utility that creates the handlers. I intend to further work on it but not having time though. sherifandroid.blogspot.com/2011/10/sax-class-generator-v10.html — Sherif elKhatib
– Sherif elKhatib, Commented May 28, 2012 at 13:24

Thilo · Accepted Answer · 2012-05-28 08:20:03Z

5

There is a constructor that takes a char[] and a range, if that is what you are looking for:

new String(ch, start, length);

answered May 28, 2012 at 8:20

Thilo

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1 Comment

Sherif elKhatib Over a year ago

Oh and needless to say, it will perform better than any.

Ram kiran Pachigolla · Accepted Answer · 2012-05-28 11:14:18Z

0

String Class has new constructor which can fulfil your requirements as follows

1) new String(char[] value)

2) new String(char[] value, int StartIndex, int numberOfCharacter)

edited May 28, 2012 at 11:14

Ram kiran Pachigolla

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answered May 28, 2012 at 8:26

Pramod Kumar

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