Typescript generic type inference based on inheritance

Question

Consider this code:

class A<T> { t?: T; }
interface B {}
class C implements A<B> {}
function f<T1 extends A<T2>, T2>(a: T1): T2 | undefined { return a.t; }

const result = f(new C());
const result2 = f(new A<B>());

It turns out that the type of result or even result2 will be unknown, while it can be inferred from the context, as C is implementing A<B> (so it can be inferred as B).

Why typescript does not do that? Is it a missing feature, a non-sound inference or is there another way to achieve desired behavior?

Ahmad · Accepted Answer · 2020-03-22 18:56:40Z

2

I could find my own answer, and I have to say TypeScript is flawless!

class A<T> { t?: T; }
interface B {}
class C extends A<B> {}

type ExtractGeneric<T> = T extends A<infer X> ? X : never;
function f<T1 extends A<T2>, T2 = ExtractGeneric<T1>>(a: T1): T2 | undefined { return a.t; }

const result = f(new C());
const result2 = f(new A<B>());

edited Mar 22, 2020 at 18:56

answered Mar 22, 2020 at 18:41

Ahmad

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2 Comments

Ahmad Over a year ago

It would be appreciated if someone could explain why this inference is not done by default.

Maciej Sikora Over a year ago

Its not done as it is just second argument. There is really revert relation, you say that the first argument depends on the second. Take into consideration that TS is not able to infer T2 as there is no demands for this type, TS does not perform relation between arguments if you dont do that explicitly. And explicitly you have relation from T1 to T2, but not in other way round

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