Characters in String : Unicode 16-bit to custom 32-bit

[Timothy Toe]

I understand that internally in Java, characters in Strings are actually Unicode characters, with each character represented with 16 bits.

So, character �L� in Unicode is 0x004C
which is also 0000 0000 0100 1100

Now, I wish to encode each of the 4 bits above into individual ASCII characters:
= 0 0 4 C
= 0x30 0x30 0x34 0x43
= 00110000 00110000 00110100 01000011

So, from the original 16-bit character in Java, I want a final 32-bit.
Eventually, I�ll need to send the final result over the network, via OutputStream/writer and socket.

Can someone help me on this ? Or give me some ideas... Thanks.

[Jesper de Jong]

Just cast the char to an integer and print out the value of the integer in hexadecimal notation.

char ch = 'L';
System.out.println(Integer.toHexString((int)ch));

(This doesn't print the leading zeroes, by the way...).

[Timothy Toe]

Your suggested code will produce "4C" which is not the final result that I require. I require each half byte - "4" and "C" - be encoded into their ASCII hex representation ..which is : 0x34 and 0x43.

Thanks.

[Jesper de Jong]

import java.io.PrintWriter;
import java.io.StringWriter;
 
public class Example
{
    public static void main(String[] args) throws Exception {
        char ch = 'L';
 
        StringWriter sw = new StringWriter();
        PrintWriter pw = new PrintWriter(sw, true);
        pw.printf("%04X", (int)ch);
 
        String s = sw.toString();
        // Now you have a String with four characters: 0, 0, 4, C
 
        for (int i = 0; i < s.length(); ++i) {
            System.out.printf("0x%02X ", (int)s.charAt(i));
        }
    }
}