CODEWITHSUNDEEP
java
Shantanu Nandan
Shantanu Nandan

Reputation: 1462

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

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

Upvotes: 1

Views: 2897

Answers (5)

Crystal Meth
Crystal Meth

Reputation: 589

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()

Upvotes: 0

mike
mike

Reputation: 5055

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

Upvotes: 1

Eldar
Eldar

Reputation: 5217

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--){

Upvotes: 2

Sam I am says Reinstate Monica
Sam I am says Reinstate Monica

Reputation: 31184

arrays are 0-indexed, change <= to <

your will get that error every time you call 

name.charAt(name.length())

Upvotes: 2

dkatzel
dkatzel

Reputation: 31648

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--){

Upvotes: 6

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