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Description
TypeScript Version: 3.4.0-dev.201xxxxx
Search Terms: Type inference, discriminated union types
Code
type AllBoxes = ABox | BBox;
export interface ICanFindBoxes {
findBoxes<T extends AllBoxes, K extends string>(boxName: K): T extends { type: K } ? T[] : never;
}
class Box<T extends string = string> implements ICanFindBoxes{
type: T;
public findBoxes<T extends AllBoxes, K extends string>(boxName: K): T extends { type: K } ? T[] : never {
throw 'argg';
}
}
export class ABox extends Box<'a'>{
}
export class BBox extends Box<'b'>{}
Expected behavior:
It compiles
Actual behavior:
Compile time error:
Property 'findBoxes' in type 'Box<T>' is not assignable to the same property in base type 'ICanFindBoxes'.
Type '<T extends AllBoxes, K extends string>(boxName: K) => T extends { type: K; } ? T[] : never' is not assignable to type '<T extends AllBoxes, K extends string>(boxName: K) => T extends { type: K; } ? T[] : never'. Two different types with this name exist, but they are unrelated.
Type '(T extends { type: K; } ? T[] : never) | ({ type: K; } & ABox)[] | ({ type: K; } & BBox)[]' is not assignable to type 'T extends { type: K; } ? T[] : never'.
Type '({ type: K; } & ABox)[]' is not assignable to type 'T extends { type: K; } ? T[] : never'.
11 public findBoxes<T extends AllBoxes, K extends string>(boxName: K): T extends { type: K } ? T[] : never {
The compiler is incorrectly identifying that the signatures are different when they are not. If I simplify the union type to only one it works (but this defeats the purpose of the signature).
This was working in < 3.4.1