Why typescript compiler is not inferring generic parameter?

Question

I have this code

type Check<F, T> = F extends string
  ? string
  : T

type Func2 = <F extends T[] | string, T>(functor: F) => Check<F, T>

const x: Func2 = (x: any) => x[0]

const y = x([1, 2])
const z = x("abc")

I am getting type unknown for y while it is perfectly fine for z.

The type inferred for calling x for y is -

const x: <number[], unknown>(functor: number[]) => unknown

Why there is unknown for T but T[] is inferred properly?

Playground

There's no inference site for T in the signature of x. F can be inferred by the type of functor, but nothing you're passing in will fix T so it falls back to unknown. And so F is only constrained to unknown[] | string. I'm not sure why you'd prefer your definition of Func2 over something like type Func2 = <F extends { [k: number]: any }>(functor: F) => F[0]. Can you explain what you're trying to do and why you're trying to do it this way? — jcalz
– jcalz, Commented Sep 26, 2020 at 19:19
@jcalz The actual code contains other types as well. Check this for the actual code. — Shivam Singla
– Shivam Singla, Commented Sep 26, 2020 at 20:14

Николай Гольцев · Accepted Answer · 2020-09-26 20:21:23Z

2

Why does this happen?

Type inference in TS doesn't work as you expect. Type variables could be inferred as I think in three ways:

based on an argument of a function that has been passed;
based on analysis of return value of a function;
based on previously inferred type variables (this rule applied only from left to right order);

That's why you have got unknown type.

Solution in your case could be:

type Check<F, T> = F extends (infer U)[]
  ? U
  : T

type Func2 = <F extends any[] | string>(functor: F) => Check<F, string>

const x: Func2 = (x: any) => x[0]

const y = x([1, 2])
const z = x("abc")

edited Sep 26, 2020 at 20:21

answered Sep 26, 2020 at 19:36

Николай Гольцев

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2 Comments

Shivam Singla Over a year ago

Thank you @Николай Гольцев The solution did work for me. But the inferred signature is correct const x: <number[], unknown>(functor: number[]) => unknown. Please check the playground

Николай Гольцев Over a year ago

Yes, it's my mistake, you're right! Answer updated.

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