Java programming exception

Question

Can any one help me? I do not understand this exception:

Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 9
    at java.lang.String.charAt(Unknown Source)
    at charPerLine.main(charPerLine.java:13)

Here's the code responsible:

import java.util.*;
public class charPerLine {

    public static void main(String[] args)throws StringIndexOutOfBoundsException {
        Scanner sc=new Scanner(System.in);
        System.out.print("Type any name:");
        String s=sc.next();
        int j= s.length()+1;
        for(int i=0;i<=j;i++){
            System.out.println(s.charAt(i));
        }
    }
}

You are accessing further than your string length. You need to post some code to let us understand why. — Tugrul Ates
– Tugrul Ates, Commented Mar 6, 2011 at 8:51

Matthew Gilliard · Accepted Answer · 2011-03-06 08:56:01Z

9

You're trying to directly access the 10th character of a string which has less than 10 characters. Something like:

"12345".charAt(9)

Remember that String indices are 0-based, hence .charAt(9) => 10th character. So "123".charAt(3) would throw, too.

edited Mar 6, 2011 at 8:56

answered Mar 6, 2011 at 8:50

Matthew Gilliard

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2 Comments

user447356 Over a year ago

Very true, common cause is such loop for (i = 0; i <= length; i++) the = will cause that error..

Matthew Gilliard Over a year ago

Which is why we have for (char c : "123".toCharArray()) {...} :)

Morten Kristensen · Accepted Answer · 2011-03-06 08:50:39Z

3

You are referencing a character at position 9, which is outside the range of the actual string. Remember to check that it's within the range [0, length[.

answered Mar 6, 2011 at 8:50

Morten Kristensen

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Comments

Quentin · Accepted Answer · 2011-03-06 11:01:50Z

3

Ah. Thanks to Greg for spotting the link to the other half of this question. He is right, it isn't fair.

Your for loop goes from 0 to whatever the length of s is plus one.

So if s was an array:

['a','b','c']

Then it would go from 0 to 4. The indexes of that array are 0, 1 and 2. So you are trying to access two point beyond the end.

You want to remove this line:

int j= s.length()+1;

And change the for loop to:

for(int i = 0; i < s.length(); i++){

edited Mar 6, 2011 at 11:01

answered Mar 6, 2011 at 9:37

Quentin

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Comments

Bernd Elkemann · Accepted Answer · 2011-03-06 09:51:57Z

2

j is s.length()+1 and then i is <=j. So: s.charAt(i) will eventually access index length()+1, which is 2 more than it is allowed to.

edited Mar 6, 2011 at 9:51

answered Mar 6, 2011 at 9:37

Bernd Elkemann

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Comments

Matthew · Accepted Answer · 2011-03-06 09:36:46Z

1

The loop should be:

for(int i=0; i < s.length(); i++){
  System.out.println(s.charAt(i));
}

The last character in the string has index:

s.length() - 1

so you need to use the guard:

i < s.length()

This way your loop terminates before i becomes s.length().

answered Mar 6, 2011 at 9:36

Matthew

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Comments

MartinStettner · Accepted Answer · 2011-03-06 09:37:29Z

1

The expression

s.charAt(i)

will throw an error, if i is larger or equal than s.length(). Try using

for (int i=0; i<s.length(); i++)
    // ...

instead.

answered Mar 6, 2011 at 9:37

MartinStettner

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Comments

mt81 · Accepted Answer · 2013-01-30 19:24:15Z

0

just replace the

int j= s.length()+1;
for(int i=0;i<=j;i++){
 ...
}

with

int j= s.length();
for(int i=0;i<j;i++){
 ...
}

:)

answered Jan 30, 2013 at 19:24

mt81

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