this post was submitted on 27 Oct 2024
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Rust

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[–] [email protected] 0 points 2 weeks ago (9 children)

(Note that I'm not the article author)

In this example, you could have just made a constant with value 0 and returned a reference to that. It would also have a 'static lifetime and there would be no leaking.

I believe the intention was to demonstrate something that works with runtime values too, and a constant 0 does not.

Btw you can just write &0 to do what you proposed, there's no need for an explicit constant/static.

[–] [email protected] 0 points 2 weeks ago (8 children)
  1. Unconvincing use-case: why is returning an Option not an option?
  2. Unconvincing objection: what concrete problems are caused by utilizing Cows?
  3. Wrong demonstrated "solution": why would one have to create a value and leak it with each call instead of using one LazyLock static?
[–] [email protected] 0 points 2 weeks ago (7 children)

I can agree that the example function is not the best usecase. But the point still stand that there's no realistic escape hatch from lifetimes and memory management in Rust.

Cow does not work when you are actually required to return a reference, e.g. if you're working with some other crate that requires that. Cow also has some more strict requirements on reborrows (i.e. you can reborrow a &'short &'long T to a &'long T, but you can only reborrow a &'short Cow<'long, T> to a &'short T).

LazyLock can solve very specific issues like static, but is not a general escape hatch. Again, the example is not the best to showcase this, but imagine if you have to perform this operation for an unknown amount of runtime values. LazyLock will only work for the very first one.

[–] [email protected] 0 points 2 weeks ago (1 children)

Cow does not work when you are actually required to return a reference

What does that even mean? Can you provide a practical example?

(I'm assuming you're familiar with Deref and autoref/autoderef behaviors in Rust.)

[–] [email protected] 0 points 2 weeks ago (1 children)

This is a kind of stupid example, but imagine you have to implement some external trait (e.g. in order to work with some dependency) with the following shape:

trait Foo {
    fn foo(&self, i: usize) -> &Bar;
}

Which is not too unreasonable, for example it's very similar to the stdlib's Index. In order to implement this trait you must return a reference, you can't return e.g. a Cow or an Arc. The fact that it takes a parameter means there might not even be one single value it has to return, so you can't even cache that inside of self with e.g. LazyLock.

Of course I'm not saying I would try to reach for an escape hatch if I had to do something like this. I would first try to see if this is an intrinsic problem in my code, or if maybe I can change the crate I'm working with to be more permissible. That is, I would try to look for proper solutions first, though Cow might not always work for that.

[–] [email protected] 0 points 2 weeks ago* (last edited 2 weeks ago)

&Bar is a reference to something. That something is either a part of self, or a part of the static context. There is no other context because there is no runtime/GC. So there is no logical not-nonsensical scenario where this would be both a valid and a limiting situation in Rust. And this is why your surface analogy to Index is invalid.

If the return value may depend on something other than self or the static context, and still need to be reference-like, then the trait definition is wrong. It should either return a Cow, or go for the obvious generalization of returning impl AsRef<Bar> values. With that generalization, references, Cow`s, and more can be returned.

There is also the possibility that the trait definition is right, and you (the implementer) are trying to break a (probably) deliberate constraint (e.g. the return value in Index being tied to &self).

I would wager a guess that what you call an escape hatchet is considered a very bad C# style anyway (or will/should be). Just like how mutable statics are considered very bad in Rust 😉

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