-s == 2⁶⁴ - s
Also, there's still a modulo in there, (-s) % s; does the compiler not use a division for that line or something?
One way of thinking about it is that in arithmetic modulo M you can add or subtract any number of M terms without changing the result.
So x and M + x and 2M + x are all the same.
So 2^64 - s is the same as 2^64 - s - 2^64 = -s.
> Also, there's still a modulo in there
Yes, but it is guarded by the earlier check, so is executed only rarely (very rarely for most moderate range values << 2^64).
IMO this is half of key insight that makes Lemire's approach fast: that the modulo can be avoided most of the time by the coarser check and for wide (e.g., 2^64) input values, this means the modulo might as well not exist, performance-wise.
E.g., if you want random values between 1 and 1,000,000 that takes ~20 bits of entropy, but if you use 25 bits and Lemire's technique you'll be running at almost full speed because of the 5 extra bits (you fall into the slow case about 3% of the time).
https://en.wikipedia.org/wiki/Two%27s_complement
Say x = (2^64 - s) in a 64 bit integer. Then (x + s) = 2^64 = 0 (as 2^64 rolls over to 0). So for the purposes of addition etc. you can say -s = 2^64 - s.