Method returning index type breaks type interference of union types

[thammi]

TypeScript Version: 3.4.0-dev.201xxxxx

Search Terms: union, index type

Code

class Test<Props extends Object> {
  constructor(public o: Props) {}
  get<K extends keyof Props>(k: K) : Props[K] { return this.o[k]; }
}

interface A {t: "A";};
interface B {t: "B";};
type u = Test<A> | Test<B>;

const tmp = new Test<B>({t: 'B'}) as u;

switch(tmp.get('t')) {
  case 'A': break;
  case 'B': break;
}

Expected behavior:

Code compiles with tmp.get('t') having type "A" | "B".

This works for:

• simple objects/interfaces
• classes
• classes with generics

Actual behavior:

Compilation fails with error:

Type '"B"' is not comparable to type '"A"'.

Type of tmp.get('t') is only "A".

Problem is not contained to string literals, but also other types. Enountered this bug using immutable-js Record.get().

Playground Link: playground

Related Issues: None found

[RyanCavanaugh]

Simplified very slightly

declare class Test<T> {
  obj: T;
  get<K extends keyof T>(k: K): T[K];
}

interface A { t: "A" }
interface B { t: "B" }

declare const tmp: Test<A> | Test<B>;

switch(tmp.get('t')) {
  case 'A': break;
  case 'B': break;
}

Not a regression - this has been the behavior since 2.1.

[RyanCavanaugh]

We don't even understand why the tmp.get call was allowed in the first place? Generic signatures should not have unified across union type boundaries... 😕

[weswigham]

I have a PR open here, but IIRC a recent change anders made should have also fixed this - I'll check.

[RyanCavanaugh]

This now works the way we want it to - the call to get is flagged as an error.

[weswigham]

@RyanCavanaugh there is some additional code in #31023 that makes the call not-an-error, as the type parameters and argument lists in use are identical, so unifying them isn't so bad.