ArrayIndexOutOfBound Error

Question

Please bear with me. I am new to java but I am struggling to code ..but I am sure I will overcome it one day so please help me ..

---

String array - Palindrome is what i am working on to build up my logical sense in java. This is what I have written so far

public class Strings1 {

public static void main(String[] args) {
    String[]  str = {"dadad", "Boomot", "mojmom"};
    String[] str1=str;
    String[] str2={};
    int i=0;
    int j=0;
    int k=0;
    char c='0';
    char e='0';
    int s = str.length;

    System.out.println("number of elements in the array" + s);
try{    
    x: for(; i<s; i++){
        y:  for( j=i,k=str[i].length()-1; k>=0; j++,k--){ //j=0; k=5

             c = (str[i].charAt(j)); // 0
             e = (str[i].charAt(k)); //5

             if(c==e)
             {
                 str2[k] = Character.toString(c);
                 continue;
             }
             else 
                 break y;

            }
            System.out.println(str[2]);
        }
}catch (Exception ex){

    ex.printStackTrace();
}

    }

}

Error :- ------------ number of elements in the array3 java.lang.ArrayIndexOutOfBoundsException: 4 at Strings1.main(Strings1.java:25)

how should I go about it? Please help.

Answer 1

k is str[i].length - 1. that means, k can be 5, and str[i] can only give you some values if i is between 0 and 2 since there are just 3 elements in str array. Hence the array out of bound exception error!

Answer 2

This:

str2[k] = Character.toString(c);

Since str2 is an empty array, every element access must fail.

What you perhaps want is to work on a single stringt to detect a palindrome (or whatever you want to do). Write a method for it. Then, write another method that handles an array of strings and uses your method for the single string. This will help you avoid nesting loops and loose control of the indexes.

Answer 3

The size of the array str2 is 0 and trying to insert the entry in array on line str2[k] = Character.toString(c); Please initialise the array str2 as String[] str2=new String[length]; and move ahead

Answer 4

 k=str[i].length()-1;

i = 0, first element of str is "dadad", length = 5, k = now 4.

str2[k] = Character.toString(c);

This array has no elements and throws ArrayIndexOutOfBoundsException

Answer 5

Not sure what you are trying to do. That inner loop looks fishy, and you have picked up a bad habits here.

Try to work without labels, eschew continue and break if you can, use locally declared variables (i.e. declare them only when you need them, not earlier) and do not try to optimize early (i.e. compute str.length() whenever you need it, it really costs you nothing).

Additionally, use assertions to verify that what you assume is indeed true. Do not forget to start you program with the -ea option to enable the asserts.

A bit of editing to show what I mean:

class Test {

    public static void main(String[] args) {
        String[] str = { "dadad", "Boomot", "mojmom" };
        String[] str1 = str;
        String[] str2 = {};

        System.out.println("number of elements in the array " + str.length);

        try {
            for (int i = 0; i < str.length; i++) {
                boolean stop = false;
                for (int j = i, k = str[i].length() - 1; k >= 0 && !stop; j++, k--) { // Work on this!!!!

                    assert 0<=j && j<str[i].length();
                    assert k<=j && k<str[i].length();

                    char c = (str[i].charAt(j)); 
                    char e = (str[i].charAt(k));

                    if (c == e) {
                        str2[k] = Character.toString(c);
                    } else {
                        // NOPE!
                        stop = true;
                    }

                }
                System.out.println(str[2]);
            }
        } catch (Exception ex) {
            ex.printStackTrace();
        }
    }
}