TypeScript：通过推断函数的参数类型来定义类型

Question

Is there any possibility to describe a type that gets inferred by a function's parameters? I need something like this:

// some fn I have no control over its params
function someFn(a: string, b?: number, c?: any): any { /* ... */ }

// my wanted type that describes the args as object-records:
const result: MyType<typeof someFn> = {
 a: 'str',
 b: 42,
 c: null
};

I have no control over the signature of the functions' parameters, so converting it into someFn(args: SomeFnArgs) and MyType<SomeFnArgs> is not an option.

I don't know if it is even possible to describe the type.

Answer 1

Yes, although not quite in that way. The names of parameters are lost when typing function signatures. The arguments type for someFn in your example is [string, number?, any] (although that example is not quite valid, since non-optional parameters may not appear after optional parameters.

You can get this type with this ( playground ):

// type Parameters<T extends Function> = T extends (...args: infer R) => any ? R : never;
// How it's implemented, Parameters is a built-in type.

function someFn(a: string, b?: number, c?: any): any { /* ... */ }

type T0 = Parameters<typeof someFn>; // type T0 = [string, (number | undefined)?, any?]

const x: T0 = ['hello', 42]; // good
const y: T0 = ['hello', 'world', '!!']; // bad