问题
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?
回答1:
Simply read it as as a byte and then convert to an 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:
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
回答2:
0xFF & number will treat the number as unsigned byte. But the resultant type is int
回答3:
You can store 8-bit in a byte If you really need to converted it to an unsigned value (and often you don't) you can use a mask
byte b = ...
int u = b & 0xFF; // unsigned 0 .. 255 value
回答4:
You can do something like this:
int value = eightBits & 0xff;
The & operator (like all integer operators in Java) up-casts eightBits to an int (by sign-extending the sign bit). Since this would turn values greater than 0x7f into negative int values, you need to then mask off all but the lowest 8 bits.
回答5:
You could simply parse it into a short or an int, which have enough range to hold all the values of an unsigned byte.
来源：https://stackoverflow.com/questions/14071361/java-how-to-parse-uint8-in-java