What does assign a generic in a trait?
from capuccino@lemmy.world to rust@programming.dev on 17 Jun 00:11
https://lemmy.world/post/31498993

Maybe the question is not well written, but it’s because I do not really know what’s happening in here. I’m learning Rust, I’m doing pretty good, but this is the second time that stomp with this.

First, I thought that only the Add trait would be enough, but the LSP keep saying me this if I do not add the “restriction”, as far as I know.

What I do not get is what <Output = T> is. I know that is using the type T, but why it is assigned to Output?

The first time that I saw something similar was in the Rust book that comes with rustup, just look at the next function signature

Thank you for you help, you are awesome.

#rust

threaded - newest

anselmschueler@ieji.de on 17 Jun 00:13 collapse

@capuccino You're not assigning anything. The Output = T is expressing a constraint. It's saying that the addition of T has to result in T again. A type can implement Add with any output type it wants.

capuccino@lemmy.world on 17 Jun 00:16 next collapse

I see. How can I define my own constraints in traits? Maybe seeing how to, I can full understand what’s happening behind

anselmschueler@ieji.de on 17 Jun 00:23 collapse

@capuccino

A trait can have an associated type. You write this "trait Foo { type Bar; }". Then, when you implement the trait, you have to specify this type, with "type Bar = Something;". This is different from the trait itself being generic, because there can only be a single associated type, so you can't implement the trait multiple times with different associated types.

anselmschueler@ieji.de on 17 Jun 00:24 collapse

@capuccino

Then, when you write somewhere that some type satisfies a trait, like "T: Foo", and Foo has an associated type, you can further specify that not only does T need to implement Foo, but that the associated type satisfies some criteria.

You can do this two ways:
"Foo<Bar = Something>" says it has to be something specific,
"Foo<Bar: Baz>" says that it has to implement another trait

capuccino@lemmy.world on 17 Jun 00:27 collapse

Amazing, so I can keep still using traits as constraints rather than types. I understand know. Hehehe, this is something that has been in my head since days ago

anselmschueler@ieji.de on 17 Jun 00:15 collapse

@capuccino An alternative is to not use T: Add<Output = T>, but to replace your usage of T, where it represents the output type, with <T as Add>::Output

capuccino@lemmy.world on 17 Jun 00:24 collapse

I found it in the book. Is in advanced chapters, near the end, dang. Thank you for your time!