Inferred return type of generic function with specialized arguments is unknown

[kol-93]

TypeScript Version: from 3.5.1 to 4.0.0-beta

Search Terms:
is:issue is:open return type generic unknown

Code

declare function f<T>(x: T): T;
declare function g(x: number): number;

type F = typeof f; // type F = <T>(x: T) => T;
type G = typeof g; // type G = (x: number) => number;

type TypedReturn<Fn extends (...args: any[]) => any, Args extends any[]> = Fn extends <A extends Args>(...args: A) => infer R ? R : never;
type TypedReturn1<Fn extends (arg: any) => any, Arg> = Fn extends <A extends Arg>(arg: A) => infer R ? R : never;

type FR = TypedReturn<F, [number]>;
// expected: type FR = number;
// actual:   type FR = unknown;

type GR = TypedReturn<G, [number]>;
// expected: type GR = number;
// actual:   type GR = number;

type FR1 = TypedReturn1<F, number>;
// expected: type FR1 = number;
// actual:   type FR1 = unknown;

type GR1 = TypedReturn1<G, number>;
// expected: type GR1 = number;
// actual:   type GR1 = number;

// These types are proofs of a bug
type F_Infer_OK = F extends (...args: [number]) => number ? true : false;
// type F_Infer_OK = true;
type F_Infer1_OK = F extends (x: number) => number ? true : false;
// type F_Infer1_OK = true;

type F_Infer_FAIL_1 = F extends (...args: [number]) => string ? true : false;
// type F_Infer_FAIL_1 = false;
type F_Infer1_FAIL_1 = F extends (x: number) => string ? true : false;
// type F_Infer1_FAIL_1 = false;

type F_Infer_FAIL_2 = F extends (...args: [string]) => number ? true : false;
// type F_Infer_FAIL_2 = false;
type F_Infer1_FAIL_2 = F extends (x: string) => number ? true : false;
// type F_Infer1_FAIL_2 = false;

Expected behavior:
See comments in Code section

Actual behavior:
See comments in Code section

Playground link

Related Issues:
No direct related issues found.

[awerlogus]

Related to #38182

[ahejlsberg]

This is effectively a duplicate of #22617. As I mention there, the fix would be to have type inference in conditional types perform instantiation of a source type in the context of a target type when the source type is a generic function type.

[typescript-bot]

This issue has been marked as a 'Duplicate' and has seen no recent activity. It has been automatically closed for house-keeping purposes.

[typescript-bot]

This issue has been marked as a 'Duplicate' and has seen no recent activity. It has been automatically closed for house-keeping purposes.