With the c++ version, you can get the same result as python like (modulus) + (value % modulus) = index -> 4 – (-1 % 4) = 3, 4 – (-2 % 4) = 2, 4 – (-3 % 4) = 1, 4 – (-4 % 4) = 0

This ought to also tie into how integer division works in those languages because I’d like to think that they all respect the identity:
(a / b) * b + a % b = a

I've always gone with (a%b+b)%b when a might be negative (and assuming b is positive). I've never looked at the assembly, but it should only be two integer instructions.

That's a surprising number of different ways that one operation is done depending on language

4:21 "If we have -7 divided by 4, is the answer -3, or is it +3?"

Aren't the choices -3 and +1? You're either 3 behind one multiple, or 1 ahead of the other.

Great explanation! I stumbled upon that Wikipedia page before but got scared away 😅

I did not know they were different. Now I need to go and experiment to see where my intuition says the answer should be.

Great topic. – If you refer to the behavior in C and C++ of the remainder relating to negative operands, the older standards specified an implementation-defined behavior. Only since C99 and C++11 the behavior is well-defined.
So, if there was an implementation which calculates -5 / 3 to be -2, then -5 % 3 needs to be 1. This would be a valid implementation-defined C++03 behavior because (a / b) * b + (a % b) = a is fulfilled.
But since the newer standards enforce – 5 / 3 to be -1 (truncation toward zero), -5 % 3 must be -2. Here, also -1 * 3 + -2 = -5 is true.

That's exactly why I added the `modulo_arithmetic` lint to Rust Clippy. It was a real pita to find that % behaves like this while chasing a bug in Python2Cpp migration I've been doing some time ago.

This doesn't account for the fact that within some languages it is also implementation defined… That even adds a bit more complexity to the behavior of the modulus operator.