Java字节数组到有符号的Int

Question

I'm trying to convert a signed int variable to a 3 byte array and backwards.

In the the function getColorint , I'm converting the int value to the byte array. That works fine!

    public byte [] getColorByte(int color1){
    byte[] color = new byte[3];
    color[2] = (byte) (color1 & 0xFF);
    color[1] = (byte) ((color1 >> 8) & 0xFF);
    color[0] = (byte) ((color1 >> 16) & 0xFF);
    return color;
    }

But if I try to convert the byte array back to the Integer with the getColorint function:

    public int getColorint(byte [] color){
    int answer = color [2];
    answer += color [1] << 8;
    answer += color [0] << 16;
    return answer;
    }

it only works for positive integer values.

Here is a screenshot during the debug:

My input int value is -16673281 but my output int value is 38143 .

Can anyone help me?

Thanks :)

Answer 1

The Color class defines methods for creating and converting color ints. Colors are represented as packed ints, made up of 4 bytes: alpha, red, green, blue. You should use it.

Answer 2

The problem here is that byte is signed. When you do int answer = color[2] with color[2] == -1 , then answer will be also -1, ie 0xffffffff, whereas you want it to be 255 (0xff). You can use Guava 's UnsignedBytes as a remedy, or simply take color[i] & 0xff which casts it to int.

Answer 3

As is Color represents in 4 bytes, you should store also an alpha channel.

From Int :

public byte [] getColorByte(int color1){
    byte[] color = new byte[4];
    for (int i = 0; i < 4; i++) {
        color [i] = (byte)(color1 >>> (i * 8));
    }
    return color;
}

To Int :

public int getColorInt(byte [] color){
    int res = ((color[0] & 0xff) << 24) | ((color[1] & 0xff) << 16) |
          ((color[2] & 0xff) << 8)  | (color[3] & 0xff);
    return res;
}