How to use `StepRng` with `choose`?

[JeanMertz]

I'm confused how I can use StepRng with choose().

I validated that my StepRng is working as expected using the same code as the example:

let sample: [u64; 3] = rng.gen();
assert_eq!(sample, [2, 3, 4]);

While the returned integers do increase by 1, as expected, I was under the impression that it would also result in a predictable, but non-constant outcome of choose(), as long as I set the increment value of the RNG to anything other than 0.

However, it looks like I'm always getting the first element of my slice back. This isn't terrible, but it would be nice if this wasn't the case, so I could rule out some situation where the RNG isn't actually used, and the business logic is wired to always return the first value of a slice, which would be undesired behaviour.

[JeanMertz]

Never mind me. StepRng isn't supposed to be used for these situations. Instead, I can use a PRNG with a constant seed when testing the implementation.

[dhardy]

Most sampling algorithms assume the RNG result is evenly distributed over the whole u32 or u64 output space; several of our implementations actually give the high bits more significance than the low bits. Using StepRng with a small step will keep the u64 counter "small" for a long time, and since our gen_range primarily uses the high bits, it could be a long time before the result changes (depending on the output range).