Hex String to byte array

[Titus Abraham]

I am puzzled,

I am trying to convert a hex string which i get from network but i am decoding it different everytime. I have no idea why. i am pasting the function which i am using .

Th Hex string is "827000000d5e"

    static public byte[] toByteArr(String no)
	{
        byte[] number = new byte[no.length()/2];
        int i;
        for (i=0; i<no.length(); i+=2)
		{
            int j = Integer.parseInt(no.substring(i,i+2), 16);
            number[i/2] = (byte)(j & 0x000000ff);
		}
        return number;
	}

    public static  byte[] HexToByte(String hex)
    {
                    int len = hex.length();
                    byte[] value= new byte[len / 2];
                    for (int i = 0; i < len; i += 2)
                      {
                        value[i / 2] = (byte) ((Character.digit(hex
                            .charAt(i), 16) << 4) + Character.digit(hex
                            .charAt(i + 1), 16));

                      }
                    return value;
    }

any idea why ... I have been puzzling on it for 2 hours now..


any help appreciated..

best regards,
Titus

[Rob Spoor]

Please Use Code Tags.

Some hints:

byte[] number = new byte[no.length()/2]; 

What if no.length() is odd? You're missing out one digit. Try the following:

int length = no.length();
if (length % 2 == 1)
{
    no = "0" + no;
    length++;
}

That also ensures your string always has an even number of characters.

number[i/2] = (byte)(j & 0x000000ff); 

The "& 0x000000ff" part is unnecessary - the cast to byte cuts off anything over 0x000000ff anyway. But it doesn't hurt though.

Your second method can be removed, because the first method does the exact same thing - take a HEX string and create a byte[]. Why use two different methods? And the first one works (after the odd-length-fix) - I know because it's identical to my own HEX-string-to-byte method

[Titus Abraham]

Thank you...