That is the long term plan. In the meantime we also have intentions of making rustc interpret what you want even when the grammar would require the disambiguation so that it can provide the appropriate suggestion in that case.

https://github.com/rust-lang/rust/pull/64700

I believe they're intentional, to disambiguate from `mod Color { struct Green; }`, and because they can be arbitrarily complex expressions.

Currently the compiler gives "not a type" when the brackets are removed; I agree that it might be to disambiguate, but it's confusing and redundant syntax. I expect { } to be used strictly for blocks in Rust.

C++ requires using the template keyword to disambiguate, i.e., one has to write `foo.template bar<22>()` as opposed to `foo.bar<22>()` (did you mean `(foo.bar < 2*2) > ()` ?).

The main reason Rust meta-programming is so much better than D and C++ is that Rust has an LL(k) grammar that's trivial to parse into ASTs that can be easily manipulated.

That feature alone is definitely worth the annoyance of having to use an editor that types `{}` for you when inputting a constant expression. It's the same with removing the `::` in `::<>` - can be done, but the costs are not worth the ergonomic improvement.

That is, I don't think the claim that the braces are redundant is correct - they are there for a reason: to keep a simple grammar, which happens to be one of the most important Rust features that everybody uses every day (every single proc macro uses this feature, that includes the `println!` in `println!("Hello World")` Rust examples).

Removing redundancy is not always a good idea, but the braces are redundant.

Can you elaborate? If you remove the braces, you break one of Rust's most important language features, an LL(2) grammar.

That is, those braces give Rust an LL(2) grammar, and therefore, are not redundant.

If you know how to preserve the LL(2) grammar in Rust while removing those braces, please explain why, since the answer would revolutionize many fields of computer science.

The braces are redundant because they do not introduce any new information, which is the definition of being redundant.

Whether the resulting grammar is or not LL(2) is orthogonal.

It's a block though, isn't it?

Color::Green is an enum variant while { Color::Green } is a (constant) block (expression) that evaluates to an instance which is used as a generic type parameter.

The difference might be more easily understandable if we look at an enum variant that holds a value, where the syntactic differences between variant and instance constructor are more clearly visible.

Color::RGB(u64, u64, u64) vs { Color::RGB(10, 20, 30) }

Hm, the RFC says the braces are needed if it is not an "identity expression" with the examples:

    const X: usize = 7;
    
    let x: RectangularArray<i32, 2, 4>;
    let y: RectangularArray<i32, X, {2 * 2}>;

So I'm still not sure if Color::Green is an identity expression, they say:

> Identity expression: An expression which cannot be evaluated further except by substituting it with names in scope. This includes all literals as well all idents

So... maybe? Depends on whether Color::Green is an ident or not. At the very least I'd expect this to work, but it doesn't yet:

    use Color::*;
    
    impl State<Green> { ... }

A name in scope is a path to a name, and Color::Green is also a path, so if impl<Green> works impl<Color::Green> would work as well.