在Java中将字节转换为二进制

Question

I am trying to convert a byte value to binary for data transfer. Basically, I am sending a value like "AC" in binary ("10101100") in a byte array where "10101100" is a single byte. I want to be able to receive this byte and convert it back into "10101100." As of now I have no success at all dont really know where to begin. Any help would be great.

edit : sorry for all the confusion I didnt realize that I forgot to add specific details.

Basically I need to use a byte array to send binary values over a socket connection. I can do that but I dont know how to convert the values and make them appear correctly. Here is an example:

I need to send the hex values ACDE48 and be able to interpret it back. According to documentation, I must convert it to binary in the following way: byte [] b={10101100,11011110,01001000}, where each place in the array can hold 2 values. I then need to convert these values back after they are sent and received. I am not sure how to go about doing this.

Answer 1

String toBinary( byte[] bytes )
{
    StringBuilder sb = new StringBuilder(bytes.length * Byte.SIZE);
    for( int i = 0; i < Byte.SIZE * bytes.length; i++ )
        sb.append((bytes[i / Byte.SIZE] << i % Byte.SIZE & 0x80) == 0 ? '0' : '1');
    return sb.toString();
}

byte[] fromBinary( String s )
{
    int sLen = s.length();
    byte[] toReturn = new byte[(sLen + Byte.SIZE - 1) / Byte.SIZE];
    char c;
    for( int i = 0; i < sLen; i++ )
        if( (c = s.charAt(i)) == '1' )
            toReturn[i / Byte.SIZE] = (byte) (toReturn[i / Byte.SIZE] | (0x80 >>> (i % Byte.SIZE)));
        else if ( c != '0' )
            throw new IllegalArgumentException();
    return toReturn;
}

There are some simpler ways to handle this also (assumes big endian).

Integer.parseInt(hex, 16);
Integer.parseInt(binary, 2);

and

Integer.toHexString(byte).subString((Integer.SIZE - Byte.SIZE) / 4);
Integer.toBinaryString(byte).substring(Integer.SIZE - Byte.SIZE);

Answer 2

For converting hexidecimal into binary, you can use BigInteger to simplify your code.

public static void sendHex(OutputStream out, String hexString) throws IOException {
    byte[] bytes = new BigInteger("0" + hexString, 16).toByteArray();
    out.write(bytes, 1, bytes.length-1);
}

public static String readHex(InputStream in, int byteCount) throws IOException {
    byte[] bytes = new byte[byteCount+1];
    bytes[0] = 1;
    new DataInputStream(in).readFully(bytes, 1, byteCount);
    return new BigInteger(0, bytes).toString().substring(1);
}

---

Bytes are sent as binary without translation. It fact its the only type which doesn't require some form of encoding. As such there is nothing to do.

To write a byte in binary

OutputStream out = ...
out.write(byteValue);

InputStream in = ...
int n = in.read();
if (n >= 0) {
   byte byteValue = (byte) n;

Answer 3

Alternative to @LINEMAN78s solution is:

public byte[] getByteByString(String byteString){
    return new BigInteger(byteString, 2).toByteArray();
}

public String getStringByByte(byte[] bytes){
    StringBuilder ret  = new StringBuilder();
    if(bytes != null){
        for (byte b : bytes) {
            ret.append(Integer.toBinaryString(b & 255 | 256).substring(1));
        }
    }
    return ret.toString();
}