StringIndexOutOfBoundsException String index out of range error

Question

I am getting this error when I enter the String "s" after entring an integer.

Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 0
    at java.lang.String.charAt(Unknown Source)
    at oneB.change(oneB.java:4)
    at oneB.main(oneB.java:26)

Following is the code: (Please note that the code is still complete and I have entered some print statements for checking)

import java.util.Scanner;
public class oneB {
    public static String change(int n, String s, String t) {

        if (s.charAt(0) == 'R') {
            return onetwo(s);
        }
        return s;
    }
    private static String onetwo(String one) {
        int c = one.indexOf('C');
        System.out.print(c);
        char[] columnarray = new char[one.length() - c - 1];
        for (int i = c + 1; i < one.length(); i++) {
            columnarray[i] = one.charAt(i);
        }
        int columnno = Integer.parseInt(new String(columnarray));
        System.out.print(columnno);
        return one;

    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System. in );
        int n = in .nextInt();
        String s = in .nextLine();
        String t = in .nextLine();
        System.out.print(change(n, s, t));
    }

}

possible duplicate of Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range stackoverflow.com/questions/13021297/… — Aniket Kulkarni
– Aniket Kulkarni, Commented Sep 20, 2013 at 12:37

Piotr Kołaczkowski · Accepted Answer · 2013-09-20 12:40:18Z

6

The call in.nextInt() leaves the endline character in the stream, so the following call to in.nextLine() results in an empty string. Then you pass an empty string to a function that references its first character and thus you get the exception.

answered Sep 20, 2013 at 12:40

Piotr Kołaczkowski

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3 Comments

AmanArora Over a year ago

Being a beginner, I barely understood what you are saying. could you suggest the edit in my code which would make it run?

Piotr Kołaczkowski Over a year ago

Add one more in.nextLine() immediately after the line n = in.nextInt();

Piotr Kołaczkowski Over a year ago

And you should also protect your change() method from receiving an empty string - so the condition should be if (!s.isEmpty() && s.charAt(0) == 'R')

Stephen C · Accepted Answer · 2013-09-20 12:51:23Z

1

Here's how I debugged it:

You are getting a StringIndexOutOfBoundsException with index zero at line 4.
That means that the String you are operating on when you call s.charAt(0) is the empty String.
That means that s = in.nextLine() is setting s to an empty String.

How can that be? Well, what is happening is that the previous nextInt() call read an integer, but it left the characters after the integer unconsumed. So your nextLine() is reading the remainder of the line (up to the end-of-line), removing the newline, and giving you the rest ... which is an empty String.

Add an extra in.readLine() call before you attempt to read the line into s.

edited Sep 20, 2013 at 12:51

answered Sep 20, 2013 at 12:42

Stephen C

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1 Comment

Stephen C Over a year ago

Like I said - "... before you attempt to read the line into s".

Sajan Chandran · Accepted Answer · 2013-09-20 13:00:23Z

1

One another solution to the problem would be instead of nextLine(), use just next().

        int n = in .nextInt();
        String s = in .next();

answered Sep 20, 2013 at 13:00

Sajan Chandran

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Comments

Kayz · Accepted Answer · 2013-09-20 12:39:47Z

0

It looks like s is an empty String "".

answered Sep 20, 2013 at 12:39

Kayz

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Comments

Prabhakaran Ramaswamy · Accepted Answer · 2013-09-20 12:47:59Z

0

   for (int i = c + 1; i < one.length(); i++) {
        columnarray[i] = one.charAt(i);   // problem is here.
    }

You need to start array index from 0. But you are starting from c + 1

    for (int i = c + 1,j=0; i < one.length(); i++,j++) {
        columnarray[j] = one.charAt(i);
     }

answered Sep 20, 2013 at 12:47

Prabhakaran Ramaswamy

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Comments

SegFault · Accepted Answer · 2013-09-20 12:48:04Z

0

The problem is that when you hit enter, your int is followed by a '\n' character. Just modify the code like this :

public static void main(String[] args) {
    Scanner in = new Scanner(System. in );
    int n = in .nextInt();
    in.nextLine(); //This line consume the /n afer nextInt
    String s = in .nextLine();
    String t = in .nextLine();
    System.out.print(change(n, s, t));
}

answered Sep 20, 2013 at 12:48

SegFault

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