java Java如何在java中解析uint8？
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Java how to parse uint8 in java?
提问by stackoverflow
I have a uint8(unsigned 8 bit integer)coming in from a UDP packet. Java only uses signed primitives. How do I parse this data structure correctly with java?
我有一个来自 UDP 数据包的uint8 （无符号 8 位整数）。Java 只使用有符号的原语。如何用java正确解析这个数据结构？
回答by Martijn Courteaux
Simply read it as as a byte and then convert to an int.
只需将其作为字节读取，然后转换为 int。
byte in = udppacket.getByte(0); // whatever goes here
int uint8 = in & 0xFF;
The bitmask is needed, because otherwise, values with bit 8 set to 1 will be converted to a negative int. Example:
需要位掩码，否则，位 8 设置为 1 的值将转换为负整数。例子：
This: 10000000
Will result in: 11111111111111111111111110000000
So when you afterwards apply the bitmask 0xFF to it, the leading 1's are getting cancelled out. For your information: 0xFF == 0b11111111
因此，当您随后对其应用位掩码 0xFF 时，前导 1 将被抵消。供你参考：0xFF == 0b11111111
回答by Prince John Wesley
0xFF & numberwill treat the number as unsigned byte. But the resultant type is int
0xFF & number将数字视为无符号字节。但结果类型是int
回答by Peter Lawrey
You can store 8-bit in a byteIf you really need to converted it to an unsigned value (and often you don't) you can use a mask
您可以将 8 位存储在 abyte如果您确实需要将其转换为无符号值（通常您不需要），您可以使用掩码
byte b = ...
int u = b & 0xFF; // unsigned 0 .. 255 value
回答by Ted Hopp
You can do something like this:
你可以这样做：
int value = eightBits & 0xff;
The &operator (like all integer operators in Java) up-casts eightBitsto an int(by sign-extending the sign bit). Since this would turn values greater than 0x7f into negative intvalues, you need to then mask off all but the lowest 8 bits.
该&运营商（如Java中的所有整数运算符）上强制转换eightBits到int（由符号位符号扩展）。由于这会将大于 0x7f 的int值转换为负值，因此您需要屏蔽除最低 8 位以外的所有位。
回答by Asik
You could simply parse it into a shortor an int, which have enough range to hold all the values of an unsigned byte.
您可以简单地将其解析为 ashort或 an int，它们有足够的范围来保存无符号字节的所有值。
