What’s the " operator" on Bash?

Question

i=1
aux=3
aux=$(($aux^$i))
echo $aux
#2

Can someone explain me this code?

Answer 1

^ is the operator "or exclusive" (bitwise XOR). Basically, it takes the binary representation of the numbers involved and applies the following rules, bit by bit:

• if both bits are equal (either 0 or both 1), the result is 0
• otherwise the result is 1

So by doing 3 ^ 1, we have:

00000011  <- 3 em binário
00000001  <- 1 em binário
--------
00000010  <- 2 em binário

In the first 6 positions, the bits are all zero, so the result of these positions will be zero. In seventh position, the first bit is 1 and the other is zero, so the result, according to the above rules, is 1. And in eighth position, both are 1, so the result is zero.

The final result is the binary representation of the number 2.

---

The expression must be between (( )), since this is the construction that allows performing arithmetic operations.

The $ is placed in front of the parentheses so that the result of the expression is assigned to the variable (aux=$(($aux^$i))). But it could also be done so:

#!/bin/bash
i=1
aux=3
((aux = $aux^$i))
echo $aux

^{See here the code running}