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This. Traits and macros are two real problems with rust. Orphan rule is one, but also const, async and unnamable types (mostly closures). These barely work with traits or do not work at all. If rust did not have closures, it'd be a lot simpler to solve these. Is it so hard to just create a normal function instead of closure?

Perhaps we need to go back to the basics a bit? What is a trait? 1. A set of functions, associated types and generic types 2. A marker/tag (e.g. Send, Sync)

Orphan rules do not seem to be problem for marker traits. Library authors must be responsible for enforcing whether their types are Send/Sync, etc or not.

As for normal traits, it's too late for rust, but I'd just limit traits to being only sets of function definitions, e.g.

trait Iterator = fn next<T>(&mut self) -> Option<T> + fn len(&self) -> usize

Then adding set operations (and, or, xor, not) for traits would be pretty easy, keeping most of power for defining generics.

More importantly traits could be just aliases and two traits with the same set of functions would be equal. This solves orphan problem - you would not need to import or export traits, it would be just normal resolution of functions. Do I call this function from crate A, or crate B? That's a solved problem.

▲ pyrale 3 days ago | parent | next [-]

The hard problem is not sharing the trait, but sharing the trait instance.

With your solution, if too modules define traits with identical type signatures but different implementations, it would be impossible for the compiler to decide which impl to use.

▲ zk4x 3 days ago | parent [-]

If there are two modules module_a and module_b, and each defines a function called foo, how does the compiler decide which foo should be used? It just checks whether you imported module_a::foo or module_b::foo.

▲ pyrale 3 days ago | parent [-]

The instance could be defined in module c.

▲ zk4x 3 days ago | parent [-]

Perhaps I should have been more clear. The point is you would not implement traits. You would just implement functions. You would not implemet traits, you would just write functions iter and len for your type.

When calling a function, compiler would check separately for existence of each function defined in the trait. That is a trait would be just like any other type alias so that you do not need to repeating complex function names everywhere:

You could write:

    trait Iterator = fn next<T>(&mut self) -> Option<T> + fn len(&self) -> usize

    fn filter(iter: impl Iterator)

but that would be just syntactic sugar for this:

    fn filter(iter: fn next<T>(&mut self) -> Option<T> + fn len(&self) -> usize)

Basically removing traits alltogether, just relying on functions.

	▲	3 days ago \| parent \| next [-]
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	▲	pyrale 3 days ago \| parent \| prev [-]
		So you define your filter function by saying it can use anything that has the next and the len functions. Cool. If a type A defined in module_a doesn’t have the functions defined in its module, should it still be filterable? If the required function is defined in a module_b should A be filterable?

▲ dgfitz 3 days ago | parent | prev [-]

> trait Iterator = fn next<T>(&mut self) -> Option<T> + fn len(&self) -> usize

This is why I don’t have interest in learning rust. When I see people write c++ code like this I wonder why they feel the need to be so clever.

That is an abysmal thing to maintain.

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marcosdumay 3 days ago | parent | next [-]

Nope, just because you lack familiarity with the symbols, it doesn't make that declaration complex.

It's not even "clever". It's completely plain, and the bare minimum and obvious definition of what an "iterator" should be.

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steveklabnik 3 days ago | parent | prev | next [-]

That isn’t real Rust code, though it’s made up of valid bits of Rust syntax.

	▲	dgfitz 3 days ago \| parent [-]
		Oh, ok that’s fair. I guess I should at least learn the grammar better before commenting more on rust.

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moomin 3 days ago | parent | prev [-]

I mean, C#'s IEnumerator is equivalent and is way more complex. You just don't need to think about it much.