What does ArrayIndexoutofBoundsException mean here?

Question

Given a string, print the number of alphabets present in the string. Input: The first line of input contains an integer T denoting the number of test cases. The description of T test cases follows.Each test case contains a single string. Output: Print the number of alphabets present in the string.

This is a question i have been trying to solve this question on eclipse but it keeps throwing ArrayIndexoutOfBoundsException in line 7 of my code. I tried understanding what i've done wrong but i have not been able to. Could some one please explain whats wrong here . I have attached the code.

public class solution {
    public static void main(String[] args){
        String s = "baibiasbfi" ;
        int count =0; 
        for(int i=0;i<=s.length();i++){
            char[] a= s.toCharArray();
            if(a[i]>='a'&& a[i]<='z'||a[i]>='A'&&a[i]<='Z')
                count++;}
    System.out.println(count);  
    }
}

Basically what ArrayIndexOutOfBoundsException means is that you're trying to access a position of your array `a` that's beyond access. In this case, your string `baibiasbfi` has 10 characters which means that you can access it from position 0 to 9. When you're doing `i <= s.length()`, and since `s.length` is equal to the size of the array which is 10, you're trying to access the position 10 of the array which is beyond reach. To solve it, remove the `=` from your condition `i <= s.length()` and you'll have your problem solved. — António Ribeiro, Feb 05 '16 at 08:00

score 0 · Answer 1 · answered Feb 05 '16 at 07:52

0

i <= s.length();

in your for loop should be:

i < s.length();

ArrayIndexOutOfBoundsException is thrown:

to indicate that an array has been accessed with an illegal index. The index is either negative or greater than or equal to the size of the array.

answered Feb 05 '16 at 07:52

thegauravmahawar

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thank you so much ............ it worked. – jenygeorge Feb 05 '16 at 09:01

score 0 · Answer 2 · answered Feb 05 '16 at 07:52

In java, arrays are from 0 to length-1. You are using your loop compare it with i<=s.length(). This means your loop is giving back the length of the array. This gives a ArrayIndexoutofBoundsException.

Replace your loop with:

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

What does ArrayIndexoutofBoundsException mean here?

2 Answers2