Java bitwise operation
java
Solution
The `0xff` is necessary because of a confluence of two factors:
- All integer types in Java are signed
- All bitwise operators promote their arguments to `int` (or `long`, if necessary) before acting.
The result is that if the high-order bit of `content[11]` was set, it will be sign-extended to a negative `int` value. You need to then `&` this with `0xff` to return it to a (positive) byte value. Otherwise when you `|` it with the result of `content[12]<<8`, the high-order byte will be all 1s.
Problem
I have this line of code ``` int b1 = 0xffff & (content[12]<<8 | 0xff & content[11]); ``` I have a bytearray (content[]) in little endian and need to recreate a 2 byte value. This code does the job just fine but prior to testing i had it written like this ``` int b1 = 0xffff & (content[12]<<8 | content[11]); ``` and the result was not right. My question is why is 0xff necessary in this scenario?