Overloaded union type does not resolve overloads

[maximilianMairinger]

Two interfaces with (at the same signatures) overloaded functions as members, combined into an union type, loose the overload information.

TypeScript Version: 3.9.0-dev.20200328

Search Terms:

Union Overload resolve interface

Code

type Wrap<T> = {wrap: T}

type A<Values extends any[]> = {
  get(): Values
  get(overload: string): Wrap<Values>
}

type B<Value> = {
  get(): Value
  get(overload: string): Wrap<Value>
}

type AB<Values extends any[]> = A<Values> | B<Values[number]> 

let ab: AB<(string | number)[]>
let res = ab.get("overload") // typeof (string | number | Wrap<(string | number)[]>)

res should be only typeof (Wrap<string | number> | Wrap<(string | number)[]>) since both subtypes (A and B) of the union type (AB) have the same overload signatures

Expected behavior:

res in the above code should be the same as A and B seperatly executed.

type AB1<Values extends any[]> = A<Values>
let ab1: AB1<(string | number)[]>
let res1 = ab1.get("overload")

type AB2<Values extends any[]> = B<Values[number]>
let ab2: AB2<(string | number)[]>
let res2 = ab2.get("overload")


let whatResShouldBe: (typeof res1) | (typeof res2)

Actual behavior:

res in the first example is typeof (string | number | Wrap<(string | number)[]>) and not as the union suggests (typeof res1) | (typeof res2) or (Wrap<string | number> | Wrap<(string | number)[]>)

Playground Link

[RyanCavanaugh]

You really, really, really want to list overloads in the other order

type A<Values extends any[]> = {
  get(overload: string): Wrap<Values>
  get(): Values
}

type B<Value> = {
  get(overload: string): Wrap<Value>
  get(): Value
}

TS proceeds with overload resolution in declared order and if you list less-specific signature first, they'll be considered as potential matches. Always have longer overloads first.