Hex to ASCII conversion shows wrong result

Question

I am using the following code to convert Hex to ASCII:

public String hexToAscii(String hex) {

    StringBuilder sb = new StringBuilder();

    StringBuilder temp = new StringBuilder();

    for(int i = 0; i < hex.length() - 1; i += 2){

        String output = hex.substring(i, (i + 2));

        int decimal = Integer.parseInt(output, 16);

        sb.append((char)decimal);

        temp.append(decimal);

    }   

    return sb.toString();

}

Input:

hexToAscii("51d37bdd871c9e1f4d5541be67a6ab625e32028744d7d4609d0c37747b40cd2d");

Expected output:

QÓ{݇žMUA¾g¦«b^2‡D×Ô`7t{@Í-

Present output:

-Í@{t7?`Ô×D?2^b«¦g¾AUM??Ý{ÓQ. 

Basically, the output I am getting is the reverse of the output that is expected. Any solutions?

Answer 1

Assuming your input string is in_str, I would use a method like this

public static byte[] decode(String in_str) {

  if (in_str != null) {

    in_str = in_str.trim();

    List<Byte> bytes = new ArrayList<Byte>();

    char[] chArr = in_str.toCharArray();


    int t = 0;

    while (t + 1 < chArr.length) {

      String token = "" + chArr[t] + chArr[t + 1];

      // This subtracts 128 from the byte value.

      int b = Byte.MIN_VALUE

          + Integer.valueOf(token, 16);

      bytes.add((byte) b);

      t += 2;

    }

    byte[] out = new byte[bytes.size()];

    for (int i = 0; i < bytes.size(); ++i) {

      out[i] = bytes.get(i);

    }

    return out;

  }

  return new byte[] {};

}

And then you could use it like this
 

new String(decode("51d37bdd871c9e1f4d5541be67a6ab625e"

  +"32028744d7d4609d0c37747b40cd2d"))

Answer 2

Hey, I use the following code:

public static void main(String[] args) {
    String hex = "51d37bdd871c9e1f4d5541be67a6ab625e32028744d7d4609d0c37747b40cd2d";
    StringBuilder output = new StringBuilder();
    for (int i = 0; i < hex.length(); i+=2) {
        String str = hex.substring(i, i+2);
        output.append((char)Integer.parseInt(str, 16));
    }
    System.out.println(output);
}