CODEWITHSUNDEEP
javastringbuffer
user3367384
user3367384

Reputation: 11

java charAt() method and surrogate

I am writing java code and want to know why the output of this code is x. I was expecting t since it is the 5th letter.

public class StringBufferDemo{
    public static void main(String args[]){
        StringBuffer sb = new StringBuffer("ttsttxctlltnt");
        System.out.println(sb.charAt(5));
    }
}

Upvotes: 0

Views: 207

Answers (3)

takendarkk
takendarkk

Reputation: 3442

It's because in java a StringBuffer object is indexed starting at 0. 1st char at position 0, 2nd char at position 1, etc...

String ------ "t t s t t x c t l l"
ArrayIndex --  0 1 2 3 4 5 6 7 8 9

Upvotes: 2

practice2perfect
practice2perfect

Reputation: 507

the index starts at 0 so the the character at 5 th posistion is x ... if u want t as the output then try the following 

 System.out.println(sb.charAt(4));

Upvotes: 1

ayniam
ayniam

Reputation: 595

The index starts from 0 and not 1. Hence in the string "ttsttxctlltnt", the character at position 5(0,1,2,3,4,5) i.e 'x' will be printed.

Upvotes: 1

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