Give the output of the following. int x=13^3&8; System.out.println(x); Please show the process.

Answer 1

Bitwise operations.
13 = 1101
3 = 0011
8 = 1000

bitwise and & has higher precedence so it is done 1st

3 & 8
0011
1000
--------
0000 = 0

next bitwise xor ^

13 ^ 0
1101
0000
--------
1101 = 13

Answer 2

The expression 13^3&8 is 13 'XOR'ed with 3 bit 'AND'ed with 8.

But bit 'AND' has higher precedence than 'XOR', so that is processed 1st:
3&8 has result 0.

Then 13^0 has result 13.

Final result is 13.