> Doesn't the TypeStable struct just hide the unknown type, not remove it?

No, of you look at the typed IR I posted, it says

    259::Union{Nothing, Some{Int64}}

and

    (259 = Base.getproperty(o, :o))

So it's inferred properly to be a small union which julia handles efficiently at runtime:

    julia> @btime type_stable()
      18.988 ns (1 allocation: 32 bytes)
    1

> Furthermore, TypeStable itself has to be on the heap, which I think means this causes two pointer lookups rather than just one.

It's on the heap because you wrote

   mutable struct

in your example and I was just mimicking you as close as possible. If you don't actually want it to be mutable, then you can just remove the keyword mutable and it'll be even faster and have no heap allocations:

    julia> @btime type_stable()
     18.426 ns (0 allocations: 0 bytes)
    1

>small union which julia handles efficiently at runtime

Oh! I didn't realize Julia already optimized small unions.

https://julialang.org/blog/2018/08/union-splitting/ https://docs.julialang.org/en/v1/devdocs/isbitsunionarrays/

I wasn't quite sure whether something like this was possible given the semantics of Julia.

Personally, I'd still prefer something more explicit, so it would be more obvious when a union is being handled efficiently. But given that small unions are optimized, I can see why Julia would make the opposite decision.

This makes me feel a lot better about writing Julia code with union types. Thanks

Yep, small union optimizations are a godsend!