<newbie> How to generate a number within a range about quickcheck

[neithernut]

I see that Arbitrary is implemented on std::ops::Range (https://docs.rs/quickcheck/1.0.3/quickcheck/trait.Arbitrary.html#impl-Arbitrary-for-Range%3CT%3E) but I can't get the syntax right.

The implementation of Arbitrary you're referring to will generate ranges themselves, not items in that range. For example, you'll get values like (0..17) rather than, say, 5.

Currently, AFAIK the only choice you (currently) have is to generate a value of an appropriate type and then restrict/transform that value to fit into your target range. For example:

let value: u32 = Arbitrary::arbitrary(g) % 10;

Plus maybe applying a few tricks to make the distribution completely uniform, but that's usually not necessary.

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Should I expect something like Arbitrary::<(0..10)>generate(g) or Arbitrary::generate<(0..10)>(g) to work? I assume (0..10) becomes an actual type?

Not sure what you did want to express. Arbitrary::generate<(0..10)>(g) would make sense if Arbitrary::generate would be a generic function (and generic over a std::ops::Range value rather than a type), but it's not. If you want to invoke a function declared in a trait (e.g. Arbitrary::arbitrary) for a specific type T implementing that trait, you just substitute that type for the trait name (e.g. u32::arbitrary()). For example, you can write the above snippet as:

let value = u32::arbitrary(g) % 10;

https://doc.rust-lang.org/book/appendix-02-operators.html#non-operator-symbols provides more details (table B-3 and B-4 at the time of writing).

[yatesco]

thanks @neithernut - my poor brain was getting befuddled and you've clarified it for me perfectly. I'm currently doing the (i32::arbitrary(g) % 9) + 1 because I want 1..=10 trick so will continue with that.