String index out of range exception and for loop is not working

Question

I am getting correct output for first for loop: for(int i=0;i<=name.length();i++) but don't know why I am not getting any output for this loop: for(int i=name.length();i>=0;i--). While executing I am getting an error saying that index out of range.

I check the error here but I didn't understand it.

public class runner {
    public static void main(String[] args) {
        String name = "java";
        System.out.println(".length method()" + name.length());// executing
                                                                // .length()
                                                                // method
        System.out.println(".charAt method()" + name.charAt(5));
        for (int i = 0; i <= name.length(); i++) {
            System.out.println(name.charAt(i));
        }
        for (int j = name.length(); j >= 0; j--) {
            System.out.println(name.charAt(j));
        }
    }
}

Output

j
a
v
a

In Java, class names should be capitalized (`Runner` instead of `runner`), and please run the formatter before pasting code. — chrylis -cautiouslyoptimistic-, Aug 14 '13 at 17:34
To simplify all answers below, indexes start at 0 meaning the maximum accessible index is length-1. Therefore, indexes must be in the range 0 <= index < length. — Josh M, Aug 14 '13 at 17:38
That could be interesting, too. http://stackoverflow.com/questions/3925130/java-how-to-get-iteratorcharacter-from-string — mike, Aug 14 '13 at 18:15

score 6 · Accepted Answer · answered Aug 14 '13 at 17:34

6

The problem is i<=name.length();

you want i<name.length(); because the length goes beyond the bounds of the wrapped char array in String.

For the same reason, you need to change the second for loop to

for(int j=name.length()-1 ;j>=0;j--){

answered Aug 14 '13 at 17:34

dkatzel

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score 2 · Answer 2 · answered Aug 14 '13 at 17:34

2

arrays are 0-indexed, change <= to <

your will get that error every time you call

name.charAt(name.length())

answered Aug 14 '13 at 17:34

Sam I am says Reinstate Monica

30,851
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100

Eldar Agalarov · Answer 3 · 2013-08-14T17:42:21.753

2

You have out of bound exception, because Java array indexing starts from zero. For example if you have array with length = 5, then index of the first element will be 0 and index of the last element will be 4 (length - 1).

Change your line

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

to

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

and line

for(int j=name.length();j>=0;j--){

to

for(int j=name.length()-1;j>=0;j--){

edited Aug 14 '13 at 17:42

answered Aug 14 '13 at 17:35

Eldar Agalarov

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score 1 · Answer 4 · answered Aug 14 '13 at 18:10

The other answers already pointed out that indexing begins at zero in java (and most other programming languages).

Since you seem to be new to java, here an example without the explicit use of indices.

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

public class Test
{

  public static void main(String[] args)
  {
    String name = "java";

    System.out.println("NORMAL ORDER\n");

    for (char c : name.toCharArray()) // for each loop
      System.out.println(c);

    System.out.println("\nREVERSE ORDER\n");

    List<Character> chars = new ArrayList<Character>(); // it's not possible to use primitive types as generic parameter

    for (char c : name.toCharArray()) // we fill the chars list
      chars.add(c); // autoboxing from char to Character

    Collections.reverse(chars);

    for (char c : chars)
      System.out.println(c);
  }
}

OUTPUT

NORMAL ORDER

j
a
v
a

REVERSE ORDER

a
v
a
j

score 0 · Answer 5 · answered Feb 25 '14 at 16:28

Always remember to check the highest index that an array can have. In your case the name.length() function returns the size of the String name.

Suppose your String is "JAVA" it contains 4 characters. So name.length() will return 4.

But notice that the indices for each character are:

'J' - 0

'A' - 1

'V' - 2

'A' - 3

When you put your index counter at 4 it tries to access something that does not exist in the bounds of the array. Hence the ArrayIndexOutOfBoundsException is raised.

For solving your problem make all name.length() calls into name.length() - 1. Or modify the for loop to not include the case when the counter equals the name.length()

String index out of range exception and for loop is not working

5 Answers5

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