Getting error in convert a string of a hex to a byte array in Java

Question

I am looking for a way to convert string that represents hex values into a byte array. This code is perfect working for even number string length but not for odd number string length. If string length is odd then I get java.lang.StringIndexOutOfBoundsException. Please help.

public byte[] hexStringToByteArray(String s) {
     byte[] data = null;
         if(! s.equals("")){    
            int len = s.length();
            data = new byte[len / 2];
            for (int i = 0; i < len; i += 2) {
                data[i / 2] = (byte) ((Character.digit(s.charAt(i), 16) << 4)
                                     + Character.digit(s.charAt(i+1), 16));
            }
         }
     return data;
}

What's wrong with String#getBytes()?

JonK
– JonK

2019-05-31 09:38:12 +00:00
Commented May 31, 2019 at 9:38 — JonK
– JonK, Commented May 31, 2019 at 9:38

DustInComp · Accepted Answer · 2019-05-31 09:39:33Z

1

Your for loop says to keep looping while i < len which allows i to be the string's last index within your loop, causing s.charAt(i+1) to access a non-existent byte.

Either change the loop condition to i < len-1 or i+1 < len.

answered May 31, 2019 at 9:39

DustInComp

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Comments

Abhishek Tiwari · Accepted Answer · 2019-05-31 09:50:24Z

0

data in case of odd number becomes of less size.

Suppose string length is 7 then data array would be of size 3,

7/2 = 3 (takes integer value)

Later in the loop you are trying to access data for 0 to 3 (which are 4 size).

Also you are accessing element in s.charAt(i+1) which is resulting larger index as your loop starts with 0 to 7. Value of i changes as :

0 2 4 6

when it is 6 then s.charAt(7) will throw StringIndexOutOfBoundsException. As string is of length 7 (index 0 to 6). 7th index is not present for string.

Solution is to reduce the no. of times loop is running by changing loop condition.

edited May 31, 2019 at 9:50

answered May 31, 2019 at 9:44

Abhishek Tiwari

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