Intersection, Union depends on type variable instantiation order

🔎 Search Terms

generic type variable intersection union order

🕗 Version & Regression Information

• This is the behavior in every version I tried, and I reviewed the FAQ for entries about Generics

⏯ Playground Link

Workbench Repro

💻 Code

type Union<A, B> = A | B
type UnionAny<T> = Union<T, any>
type UnionUnknown<T> = Union<unknown, T>

// these should all be the same type but are respectively: any, any, unknown 
type UA0 = Union<unknown, any>
//   ^?
type UA1 = UnionAny<unknown>
//   ^?
type UA2 = UnionUnknown<any>
//   ^?

type Intersect<A, B> = A & B
type IntersectAny<T> = Intersect<T, any>
type IntersectNever<T> = Intersect<never, T>

// these should all be the same type but are respectively: never, any, never
type AN0 = Intersect<never, any>
//   ^?
type AN1 = IntersectAny<never>
//   ^?
type AN2 = IntersectNever<any>
//   ^?

🙁 Actual behavior

The types UA0, UA1 are resolved as any and UA2 is resolved as unknown. These all map down to the same union of unknown and any, so should all be the same type.

The types AN0, AN2 are resolved as never and AN1 resolved as any. These all map down to be the intersection of any and never, so should all be the same type.

🙂 Expected behavior

I expect UA0, UA1 and UA2 to resolve as the same type (probably any, but preferably unknown)
I expect AN0, AN1 and AN2 to resolve as the same type (probably never)

Additional information about the issue

No response

Beginne mit dem Workbench Repro und vergleiche die aufgelösten Typen von UA0, UA1, UA2, AN0, AN1 und AN2. Verfolge das Verhalten des Type-Checkers bei generischer Instanziierung, Vereinigungen und Schnittmengen; abgeschlossen ist die Aufgabe, wenn äquivalente Ausdrücke konsistent aufgelöst werden und die erwarteten Typen des Repros gelten.