从 2 或 4 字节转换为有符号/无符号 short/int
我必须将字节转换为有符号/无符号整数或短整数.
I have to convert bytes to signed/unsigned int or short.
以下方法正确吗?哪些已签名，哪些未签名?
The methods below are correct? Which is signed and which unsigned?
字节顺序:LITTLE_ENDIAN
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));
}
和
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));
}
我仅对此转换表单感兴趣.谢谢！
I'm interested only in this conversion forms. Thanks!
推荐答案
每对的第一个方法(convertXXXToInt1())被签名，第二个方法(convertXXXToInt2()) 未签名.
The first method (convertXXXToInt1()) of each pair is signed, the second (convertXXXToInt2()) is unsigned.
然而，Java int 总是有符号的，所以如果设置了 b4 的最高位，convertFourBytesToInt2() 的结果将是负数，即使这应该是未签名"版本.
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.
假设一个byte值，b2是-1，或者十六进制的0xFF.<< 运算符将导致它被提升"为值为 -1 或 0xFFFFFFFF 的 int 类型.移位8位后为0xFFFFFF00，移位24字节后为0xFF000000.
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.
但是，如果您应用按位 & 运算符，则高位将设置为零.这将丢弃符号信息.以下是这两个案例的第一步，更详细地制定了.
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.
签名:
byte b2 = -1; // 0xFF
int i2 = b2; // 0xFFFFFFFF
int n = i2 << 8; // 0x0xFFFFFF00
无符号:
byte b2 = -1; // 0xFF
int i2 = b2 & 0xFF; // 0x000000FF
int n = i2 << 8; // 0x0000FF00
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