Generic types can't have some types inferred if they are derived from another generic which doesn't necessarily use them.

[SephReed]

Bug Report

🔎 Search Terms

Generics, infer, derived, unused, doesn't work

⏯ Playground Link

Playground link with relevant code

💻 Code

type A<T, Y, Z> = () => true;  // this does not work
// type A<T, Y, Z> = () => T;  // this works

type InferT<I> = I extends A<infer T, any, any> ? T : never;

// typeof directInfer is "true"
let directInfer: InferT<A<true, null, null>>; // this works either way


type B<T> = A<T, null, null>;
const test: B<true> = () => true;

// typeof indirectInfer is "uknown"
// should be of type "true", and it works if you uncomment the line above
let indirectInfer: InferT<typeof test>;

🙁 Actual behavior

• If you make a generic A which does not necessarily use all of its arguments, then instantiate a variable of that type, you can still use infer to pull out every generic argument from it.

• If you then make a new generic B which is a version of A with some fields pre-filled, then use the same inference function as for A, it will no longer work on generic arguments which are not used in A.

• If you then change A to use all of its generic arguments, inferring from B will now work.

🙂 Expected behavior

It shouldn't matter whether or not the base generic A uses a generic field. If it can be inferred one way, it should be able to be inferred either way.

[jcalz]

This is working as intended; see this FAQ entry. TypeScript's type system is structural and not nominal; if a type doesn't depend structurally on T, then in principle you can't infer T from it. (In practice, sometimes the compiler can do this, but it really shouldn't be relied upon).

See #40796, #26815

[SephReed]

Ah. I suppose that makes sense. Luckily I found a way of getting around my issue so I guess I'll close this.