Type parameter that allows for the retrieval of a function parameter's type based on specific conditions

Question

Type parameter that allows for the retrieval of a function parameter's type based on specific conditions

There is a custom type I've developed to extract the type of the second parameter in a function:

type SecondParam<T> = T extends (a: any, b: infer R) => any
  ? R
  : never;

For the most part, it functions correctly:

type T1 = SecondParam<(x: any, y: number) => void>; // number
type T2 = SecondParam<(x: any, y: { x: string[] }) => void>; // { x: string[] }

~~However, when the second parameter is absent, I want it to return void instead of an empty object:~~

~~type T3 = SecondParam<(x:any) => any> // currently {}, but I prefer void~~

I'm utilizing the SecondParam type to specify another function's type:

type F<T> = (p: SecondParam<T>) => void;
type F1 = F<(x: any, y: number) => any>; // (p: number) => void
type F2 = F<(x: any, y: { x: string[] }) => any>; // (p: { x: string[] }) => void
type F3 = F<(x: any) => any>; // currently (p: {}) => void, but I prefer () => void

Unfortunately, it doesn't handle cases where the second parameter is missing as desired.

typescript conditional-types

Answer 1

Answer №1

The condition

(a: any, b: infer R) => any ? .. : ..

will always evaluate to true because omitting parameters does not result in compile errors as shown in this example:

const abc: (one: string, two: string, three: string) => void = (onlyFirst: string) => { };

If you actually want () => void as the type, then you need to introduce a new type condition for the SecondParam type like this:

type SecondParam<T> = T extends (a: any, b: infer R) => any
  ? R
  : never;

type F<T> = T extends (a: any) => any
  ? () => void
  : (p: SecondParam<T>) => void;

type F3 = F<(x: any) => any>; // is () => void

If you require the type never, which cannot be assigned directly, you can modify the SecondParam type like this:

type SecondParam<T> = T extends (a: any) => any
  ? never
  : T extends (a: any, b: infer R) => any
    ? R
    : never;

// Since it expects only one argument due to the never type:
const f: F3 = () => { };
f(); // expects 1 argument

Answer 2

The condition

(a: any, b: infer R) => any ? .. : ..

will always evaluate to true because omitting parameters does not result in compile errors as shown in this example:

const abc: (one: string, two: string, three: string) => void = (onlyFirst: string) => { };

If you actually want () => void as the type, then you need to introduce a new type condition for the SecondParam type like this:

type SecondParam<T> = T extends (a: any, b: infer R) => any
  ? R
  : never;

type F<T> = T extends (a: any) => any
  ? () => void
  : (p: SecondParam<T>) => void;

type F3 = F<(x: any) => any>; // is () => void

If you require the type never, which cannot be assigned directly, you can modify the SecondParam type like this:

type SecondParam<T> = T extends (a: any) => any
  ? never
  : T extends (a: any, b: infer R) => any
    ? R
    : never;

// Since it expects only one argument due to the never type:
const f: F3 = () => { };
f(); // expects 1 argument

Type parameter that allows for the retrieval of a function parameter's type based on specific conditions

Answer №1

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