java 从 2 或 4 个字节转换为有符号/无符号短/整数
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Convert from 2 or 4 bytes to signed/unsigned short/int
提问by blackwolf
I have to convert bytes to signed/unsigned int or short.
我必须将字节转换为有符号/无符号整数或短整数。
The methods below are correct? Which is signed and which unsigned?
下面的方法对吗？哪个是签名的，哪个是未签名的？
Byte order: LITTLE_ENDIAN
字节顺序：LITTLE_ENDIAN
public static int convertTwoBytesToInt1(byte b1, byte b2) {
return (int) ((b2 << 8) | (b1 & 0xFF));
}
VS.
对比。
public static int convertTwoBytesToInt2(byte b1, byte b2) {
return (int) (( (b2 & 0xFF) << 8) | (b1 & 0xFF));
}
and
和
public static int convertFourBytesToInt1(byte b1, byte b2, byte b3, byte b4){
return (int) ((b4<<24)+(b3<<16)+(b2<<8)+b1);
}
VS.
对比。
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 onlyin this conversion forms. Thanks!
我只对这种转换形式感兴趣。谢谢！
回答by erickson
The first method (convertXXXToInt1()) of each pair is signed, the second (convertXXXToInt2()) is unsigned.
convertXXXToInt1()每对的第一个方法 ( ) 是有符号的，第二个 ( convertXXXToInt2()) 是无符号的。
However, Java intis always signed, so if the highest bit of b4is set, the result of convertFourBytesToInt2()will be negative, even though this is supposed to be the "unsigned" version.
但是，Javaint始终是有符号的，因此如果设置了最高位b4，则结果convertFourBytesToInt2()将为负，即使这应该是“无符号”版本。
Suppose a bytevalue, b2is -1, or 0xFF in hexadecimal. The <<operator will cause it to be "promoted" to an inttype 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.
假设一个byte值为b2-1 或 0xFF 的十六进制值。所述<<操作者将导致其被“提升”到一个int为-1，或0xFFFFFFFF值类型。移位8位后为0xFFFFFF00，移位24个字节后为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.
但是，如果应用按位运算&符，高位将设置为零。这会丢弃标志信息。以下是这两个案例的第一步，更详细地制定了。
Signed:
签：
byte b2 = -1; // 0xFF
int i2 = b2; // 0xFFFFFFFF
int n = i2 << 8; // 0x0xFFFFFF00
Unsigned:
未签名：
byte b2 = -1; // 0xFF
int i2 = b2 & 0xFF; // 0x000000FF
int n = i2 << 8; // 0x0000FF00
回答by dmolony
There is a problem with the 4-byte unsigned conversion, because it doesn't fit into an int. The routines below work correctly.
4 字节无符号转换存在问题，因为它不适合 int。下面的例程正常工作。
public class IntegerConversion
{
public static int convertTwoBytesToInt1 (byte b1, byte b2) // signed
{
return (b2 << 8) | (b1 & 0xFF);
}
public static int convertFourBytesToInt1 (byte b1, byte b2, byte b3, byte b4)
{
return (b4 << 24) | (b3 & 0xFF) << 16 | (b2 & 0xFF) << 8 | (b1 & 0xFF);
}
public static int convertTwoBytesToInt2 (byte b1, byte b2) // unsigned
{
return (b2 & 0xFF) << 8 | (b1 & 0xFF);
}
public static long convertFourBytesToInt2 (byte b1, byte b2, byte b3, byte b4)
{
return (long) (b4 & 0xFF) << 24 | (b3 & 0xFF) << 16 | (b2 & 0xFF) << 8 | (b1 & 0xFF);
}
public static void main (String[] args)
{
byte b1 = (byte) 0xFF;
byte b2 = (byte) 0xFF;
byte b3 = (byte) 0xFF;
byte b4 = (byte) 0xFF;
System.out.printf ("%,14d%n", convertTwoBytesToInt1 (b1, b2));
System.out.printf ("%,14d%n", convertTwoBytesToInt2 (b1, b2));
System.out.printf ("%,14d%n", convertFourBytesToInt1 (b1, b2, b3, b4));
System.out.printf ("%,14d%n", convertFourBytesToInt2 (b1, b2, b3, b4));
}
}
