Error in String comparison java program

Question

I have written the below program to compare two strings, but I am getting ArrayIndexOutOfBoundException. I don't understand why. Please review the code below and help me. Thanks in advance:

public class IPrepTest4 {

    static String s1 = "Java";
    static String s2 = "Jav";
    static String s3 = "Java";
    static boolean b = false;

    static char[] arr1 = s1.toCharArray();
    static char[] arr2 = s2.toCharArray();
    static char[] arr3 = s3.toCharArray();

    public static void main(String[] args)
    {   
        //compareString(s1,s2);
        compareString(s1,s3);
    }

    public static void compareString(String s1, String s2)
    {   
        if (s1.length() == s2.length()) {
            int i = 0;
            int j = 0;
            while (i<s1.length() && j<s2.length()) {
                b = (arr1[i] == arr2[j]);
                i++;j++;
            }
            if (b)
                System.out.println("String s1: " + s1 + " and String s2: " + s2 + " are equal");

        } else {
            System.out.println("String s1: " + s1 + " and String s2: " + s2 + " are not equal");
        }
    }
}

I don't know man but I think your code is a bit too complicated for no reason.Besides, I don't get the error you reported when I run the code. And what if s1.length does not equal s2.length? What happens then? what becomes of i and j? — Ojonugwa Jude Ochalifu
– Ojonugwa Jude Ochalifu, Commented Dec 13, 2014 at 20:31
If s1.length doesn't equals to s2.length then it should got to else block and then there is no need of i and j. I am still getting the same Exception, if you can give me any idea to change the method implementation then it would be very helpful — Bijay Singh
– Bijay Singh, Commented Dec 14, 2014 at 5:25
If s1.length doesn't equal s2.length, it doesn't get to the else block.The else block in your code belongs to another IF statement. — Ojonugwa Jude Ochalifu
– Ojonugwa Jude Ochalifu, Commented Dec 14, 2014 at 10:51

Chris Long · Accepted Answer · 2014-12-13 22:57:13Z

1

Your main() function passes in s1 and s3 , but your function has a hard-coded reference to arr2, which only has three elements. Any real implementation of a function like yours should create the char arrays dynamically, assuming that it's necessary to create those arrays at all.

answered Dec 13, 2014 at 22:57

Chris Long

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3 Comments

Bijay Singh Over a year ago

Thanks @Chris Long I found out the problem and rectified it. In the meantime can you please suggest what can I do if I have to compare to strings of different length, in my case s1 and s2, should I create any char array or not?

Ojonugwa Jude Ochalifu Over a year ago

You do not need to create a char array. s1.length is 4.You do not need a char array to count the length of a string.

Chris Long Over a year ago

You don't ever need to create the arrays, because you can use the charAt method to access a String's characters, but then I assume that this is some sort of homework assignment, because otherwise you'd be using String.equals() instead of writing your own. If you're just testing for equality / non-equality, your initial comparison of the String's lengths is valid, and you only need to compare the characters of strings that are equal lengths. If you want to do a more complex comparison (deciding which sorts first alphabetically, for example), then you need to specify that.

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