How does this code check for parity of a number? (even or odd)

Question

I came across this piece of code on reddit

1 - ((num & 1) << 1) as i32

This code returns 1 for even numbers and -1 for odd numbers.

It takes less instructions than other ways of calculating the same thing, and is presumably, quite fast. So, how does it work? (A step-by-step breakdown would be helpful)

Note: I found What is the fastest way to find if a number is even or odd?, but don't understand how that works either.

Answer 1

Let's break this down from the inside out.

1. num & 1

This "masks" all but the least significant bit using a bitwise and. Since the least significant bit is the "ones" place, it will evaluate to 1 if the number is odd or 0 if the number is even.

2. (result1) << 1

This bitshifts that left by 1. This has the effect of multiplying by two. If num was odd, this will evaluate to 2 or still 0 if num was even. (0 * 2 = 0)

3. (result2) as i32

This casts the resulting unsigned integer (2 or 0) into a signed integer, allowing us to subtract it in the next operation. This is only for the compiler, it has no effect on the value in memory.

4. 1 - result3

This subtracts the previous number from 1. If num was even, we get 1 - 0 which results in a final answer of 1. If num was odd, we get 1 - 2 which results in a final answer of -1.