CODEWITHSUNDEEP
typescript
Supremus
Supremus

Reputation: 151

Get type of array values passed to function

I have function where i accept array of string literals, and then infer type of second argument based on given array. As the simplest example let's just make second argument be one of array values

type Foo = (arr: any[], bar: typeof arr[number]) => any

As you can probably guess bar type is actually any since it is any[] in type declaration even if i provide something like 

fooFunc([const1, const2], somethingElse)

Typescript will allow it. I understand that you cannot make const arguments at least for now, but is there any workaround to make this work?

Simple changing arr type to string[] won't work

    const Codes = {
  "NOT_FOUND": "404",
  "FORBIDDEN": "403",
  "OK": "200"
   } as const

type Foo = (arr: string[], bar: typeof arr[number]) => any
const fooFunc: Foo = (arr, bar) => null;
fooFunc([Codes.FORBIDDEN, Codes.NOT_FOUND], "random string") //"random string" is string so it's completely ok

Changing arr type to Codes[] won't do the trick either, since it will allow Codes.OK which is not in our array.

Upvotes: 1

Views: 70

Answers (1)

Aplet123
Aplet123

Reputation: 35512

Some good ol' typescript hacking will solve your issues.

const Codes = {
    "NOT_FOUND": "404",
    "FORBIDDEN": "403",
    "OK": "200"
} as const;
type Foo = <T extends typeof Codes[keyof typeof Codes], K extends T>(arr: T[], bar: K) => any;
const fooFunc: Foo = (arr, bar) => null;
fooFunc([Codes.NOT_FOUND, Codes.FORBIDDEN], Codes.NOT_FOUND); // allowed
fooFunc([Codes.NOT_FOUND, Codes.FORBIDDEN], Codes.FORBIDDEN); // allowed
fooFunc([Codes.NOT_FOUND, Codes.FORBIDDEN], Codes.OK); // errors
fooFunc(["123"], "123"); // errors

See it in action here.

Upvotes: 2

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