Convert from 2 or 4 bytes to signed/unsigned short/int
byte, int, java, type-conversion
Solution
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
Problem
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!