![]() Int i = b & 0xFF // an int representing the same value Int i = 200 // some value between 0 and 255īyte b = (byte) i // 8 bits representing that valueīyte b = 123 // 8 bits representing a value between 0 and 255 Here are a couple of useful conversions / manipulations for your convenience: Conversions to / from int // From int to unsigned byte If a method accepts a byte, then that method accepts a value between −128 and 127 unless explicitly stated otherwise. Just keep in mind that there's nothing you can do to force your interpretation upon someone else's method. That being said, nothing prevents you from viewing a byte simply as 8 bits and interpret those bits as a value between 0 and 255. For instance, if a byte is cast to an int Java will interpret the first bit as the sign and use sign extension. A byte according to the language spec represents a value between −128 - 127. The Java Language does not provide anything like the unsigned keyword. (unsignedByte) // "200"Ĭomplete guide for working with unsigned bytes in Java: Unsigned byte in Java `short`, `int`, `long` and even `char`, but during bitwise operations * You could use any integer type longer than `byte` for the `unsignedByte` variable, * and than you use the `unsignedByte` variable in all your bitwise * unsigned byte and the advantage is you apply `0xFF` only once * You would typically do this *within* the method that expected an * b) were set to 1s because of "sign extension" during the upcasting * So the `b & 0xFF` is equivalent with `((int) b) & 0xFF`. * just before evaluating the bitwise AND operator. * a) were added during upcasting to int which took place silently * zero all the 24 most significant bits that: * Will print a positive int 200 because bitwise AND with 0xFF will * But you could still choose to interpret this as +200. * so called "sign extension" which yields those bits: * Will print a negative int -56 because upcasting byte to int does There is no magic flag to say "this is signed" or "this is unsigned".Īs primitives are signed the Java compiler will prevent you from assigning a value higher than +127 to a byte (or lower than -128). The fact that primitives are signed in Java is irrelevant to how they're represented in memory / transit - a byte is merely 8 bits and whether you interpret that as a signed range or not is up to you.
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