How to deal with bitwise operations in Java?

Question

Java

int a = 0xFF0;
int b = 0xF0F;
int c = 0x0FF;
c = a = (a^b | a & b)^c;
System.out.println((c | a) ^ (c & a));
==
Gives a result of 0
, write me a step-by-step solution plan
as you would decide on paper, otherwise nothing is clear no matter how hard you try , in theory you need to translate from 16 to 2, but it should be a simpler way for
a wildly long time . I ask for your help

Answer 1

use >>> or <<< ?

Answer 2

"I'm lazy and don't want to do anything. Do it for me." So you won't get anywhere.
Convert from 16 -> 2 just a couple of seconds or banned in Google? And there, google the principle of operation and the priority of logical operations should not be a problem

Answer 3

16 -> 2
This is a right shift of 3 bits, the fastest operation ever

Answer 4

"in theory, it is necessary to translate from 16 to 2, but it's wildly long"
any sane calculator will do it for you within a couple of milliseconds

Answer 5

in theory, it is necessary to translate from 16 to 2, but this is wildly long

0xFF0 = similar numbers are translated simply - F is 1111.
0xFF0 == 111111110000

Answer 6

There is no need to convert anything here, here you need to look at the boolean logic.
Namely, on (c | a) ^ (c & a) . If c=a, this expression always gives 0. Since if c=a=1 then the left will be 1 and the right will be 1, then the xor operation will return 0. For 0, the left and right will be 0, xor is also 0)