Java binary literals - Value -128 for byte

问题

Since SE 7 Java allows to specify values as binary literal. The documentation tells me 'byte' is a type that can hold 8 Bit of information, the values -128 to 127.

Now i dont know why but i cannot define 8 bits but only 7 if i try to assign a binary literal to a byte in Java as follows:

byte b = 0b000_0000;    //solves to the value 0
byte b1 = 0b000_0001;   //solves to the value 1
byte b3 = 0b000_0010;   //solves to the value 2
byte b4 = 0b000_0011;   //solves to the value 3     

And so on till we get to the last few possibilitys using those 7 bits:

byte b5 = 0b011_1111;   //solves to the value 63
byte b6 = 0b111_1111;   //solves to the value 127

If i want to make it negative numbers i have to add a leading - in front like this:

byte b7 = -0b111_1111;   //solves to the value -127

Now half of the problem i have is that i use just 7 bits to describe what they tell me is a 8 bit data type. Second half is that they dont seem to be threaded as twos complement unless using a 32bit int type where i can define all of the 32 bits ("sign indicator bit" included).

Now when i search on how to display the in-range number -128 i was told to do it this way without any further explanation:

byte b8 = 0b1111_1111_1111_1111_1111_1111_1000_0000;

I can clearly see that the last 8 Bit (1000 0000) do represent -128 in twos compelment using 8 Bit, still i never was confused more and try to ask my questions:

• Isnt the above 32 bit long nr a 32 Bit (java-) int value?
• Why can i assign a 32 bit value to a 8 bit (java-) byte type?

Or in general: Why do i have to assign it this way?

Any links/ informations about this would be great! Thank you for the time you took to read this as well as any further information in advance.

Regards Jan

回答1:

According to the Java Specification,

http://docs.oracle.com/javase/specs/jls/se7/html/jls-3.html#jls-3.10.1

all your declarations (b, b1,..., and b8) use int literals, even when they would fit in a byte. There's no byte literal in Java, you can only use an int to initialize a byte.

I did some tests and byte neg128 = -0b1000_0000; works fine. 0b1000_0000 is 128, so you just need to put a - sign before it. Notice that that 1 is not a sign bit at all (don't think about 8-bit bytes, think about 32-bit ints converted to bytes). So if you want to specify the sign bit you need to write all 32 bits, as you have demonstrated.

So byte b8 = 0b1000_0000; is an error just like byte b8 = 128; is an error (+128 does not fit in a byte). You can also force the conversion with a cast:

byte b = (byte) 0b1000_0000; or byte b = (byte) 128;

The cast tells the compiler that you know 128 does not fit in a byte and the bit-pattern will be reinterpreted as -128.

来源：https://stackoverflow.com/questions/16775169/java-binary-literals-value-128-for-byte