Byte initialization using 0xFF instead of 0xFFFFFFFF

Question

I have a question about initializing byte in java, I want to initialize a byte value allBitsOne and all bits of it are 1:

Method 1:

byte allBitsOne = 0xFF;

Wrong, it says that 0xFF is a integer type and over the range of byte, so i do it like below

Method 2:

byte allBitsOne = (byte)0xFF;

Works fine.

Method 3:

byte allBitsOne = 0xFFFFFFFF;

It works fine as well, but if 0xFF exceeds the range of a byte, why doesn't 0xFFFFFFFF?

Thank you all, I found this: link

score 4 · Accepted Answer · answered Dec 22 '13 at 17:45

4

byte is a signed integer type, going from -128 to 127.

0xFF is 255, so it's larger than 127.

0xFFFFFFFF is -1, so it's within the bounds of the byte type.

answered Dec 22 '13 at 17:45

JB Nizet

1

+1 There is no `byte` or `short` constants, only `int` (and `long`, `float`, `double`) constants. – Peter Lawrey Dec 22 '13 at 17:55

score 1 · Answer 2 · answered Dec 22 '13 at 17:44

1

literal integers in Java are signed 32 bit numbers, so:

0xff is an integer type which equals to 255, which is over the limit for byte.

0xffffffff is an integer type which equals to -1, which is not over the limit for byte.

answered Dec 22 '13 at 17:44

MByD

score 1 · Answer 3 · answered Dec 22 '13 at 17:50

1

the byte-variable in the java in the java can hold the values from -128 to 127. if you want to set all the bit to 1. then u can store the -128 to it.

answered Dec 22 '13 at 17:50

3 Answers3