Not actually unconstrained const generic

help
1

I would like to be able to use (but not have to name) a const generic in a generic type, such as:

struct Generic<T>(::core::marker::PhantomData<T>);

trait NameConst<const N: usize> {}

impl<T: NameConst<N>, const N: usize> Generic<T> {}

The compiler error is pretty clear about the issue, but it is disappointing:

"proving the result of expressions other than the parameter are unique is not supported"

And here's a playground link, in case I'm missing something.

2

Unless I’m misunderstanding what you’re trying to do, you’d need to write it more like this, which also isn’t supported— likely because the only current way for a type to satisfy the bound would be ~264 explicit impls:

impl<T> Generic<T> where T: for<const N:usize> NameConst<N> {}
3

Well, yeah, more or less. To start with I do have a simple variety:

struct L<const N: usize>;

impl<const N: usize> NameConst<N> for L<N> {}

But also, more complex versions:

struct Complex<const N: usize, const A: bool, /* ... */, const F: bool>;

impl<const N: usize, const A: bool, /* ... */, const F: bool> 
NameConst<N> for Complex<N, A, B, C, D, E, F>
where
    // Lots of bounds to map bools to N
{}

And I don't want to have to do:

impl<const N: usize> Generic<L<N>> {}
impl<const N: usize, const A: bool, /* ... */, const F: bool> 
Generic<Complex<N, A, B, C, D, E, F>> 
where
    // So many bounds
{}
4

What about:

trait NameConst {
    const N:usize;
}

impl<T:NameConst> Generic for T {
    // T::N can be used here…
}
5

Oh, there's the issue. The const of the trait doesn't guarantee it's implemented for an equivalently unique type:

struct NoConst;

trait ConstTrait<const N: usize> {}

impl<const N: usize> ConstTrait<N> for NoConst {}
6

And that gave a clue to the solution! [playground]

Just need to name it as an associated type, not a generic:

impl<T: NameType<Assoc = L<N>>, const N: usize> Final for T {}