Question

I have to convert bytes to signed/unsigned int or short.

The methods below are correct? Which is signed and which unsigned?

Byte order: LITTLE_ENDIAN

public static int convertTwoBytesToInt1(byte b1, byte b2) {
    return (int) ((b2 << 8) | (b1 & 0xFF));
}

VS.

public static int convertTwoBytesToInt2(byte b1, byte b2) {
    return (int) (( (b2 & 0xFF) << 8) | (b1 & 0xFF));
}

and

public static int  convertFourBytesToInt1(byte b1, byte b2, byte b3, byte b4){
    return (int) ((b4<<24)+(b3<<16)+(b2<<8)+b1);
}

VS.

public static int  convertFourBytesToInt2(byte b1, byte b2, byte b3, byte b4){
    return (int) (( (b4 & 0xFF) << 24) | ((b3 & 0xFF) << 16) | ((b2 & 0xFF) << 8) | (b1 & 0xFF));
}

I’m interested only in this conversion forms. Thanks!

Answer 1

The first method (convertXXXToInt1()) of each pair is signed, the second (convertXXXToInt2()) is unsigned.

However, Java int is always signed, so if the highest bit of b4 is set, the result of convertFourBytesToInt2() will be negative, even though this is supposed to be the “unsigned” version.

Suppose a byte value, b2 is -1, or 0xFF in hexadecimal. The << operator will cause it to be “promoted” to an int type with a value of -1, or 0xFFFFFFFF. After the shift of 8 bits, it will be 0xFFFFFF00, and after a shift of 24 bytes, it will be 0xFF000000.

However, if you apply the bitwise & operator, the higher-order bits will be set to zero. This discards the sign information. Here are the first steps of the two cases, worked out in more detail.

Signed:

byte b2 = -1; // 0xFF
int i2 = b2; // 0xFFFFFFFF
int n = i2 << 8; // 0x0xFFFFFF00

Unsigned:

byte b2 = -1; // 0xFF
int i2 = b2 & 0xFF; // 0x000000FF
int n = i2 << 8; // 0x0000FF00