Typescript instantiate string union type

Question

So I have a string union type

type S = 'a' | 'b' | 'c';

I want to initialize a value of type S, while asserting that is of type S. With a variable, it's very easy:

const s1: S = 'a';
const s2: S = 'z'; // error

However, I just want to instantiate the value itself, along the lines of 'a' as S but checking that 'a' is of type S.

The exact use case is that I'm using a library that provides a function f(s: string), but I want to ensure that when I call it, I only call it with the strings I deem ok. Since [afaik] you can't narrow the signature from outside the library, I was thinking of doing something like f('a' as S) every time I used f.

Defining my own function g = (s: S) => f(s) isn't a great option because actually f's signature looks like f(s: string, t: T) where T is a complex, library defined type that I'm not sure is even exported.

What would be the best way to accomplish this?

Ori Drori · Accepted Answer · 2022-03-12 00:20:34Z

2

Wrap the external function with your own using the string type (TS playground):

const externalFn = <T>(s: string, t: T) => t

type S = 'a' | 'b' | 'c';

const yourFn = (s: S, t: Parameters<typeof externalFn>[1]) => externalFn(s, t)

If you need to handle a function with more than one argument, you can use the idea in this answer (TS playground):

type DropFirst<T extends unknown[]> = T extends [any, ...infer U] ? U : never

const externalFn = <T>(s: string, t: T) => t

type S = 'a' | 'b' | 'c';

const yourFn = (s: S, ...rest: DropFirst<Parameters<typeof externalFn>>) => externalFn(s, ...rest)

edited Mar 12, 2022 at 0:20

answered Mar 12, 2022 at 0:08

Ori Drori

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5 Comments

Simon Berens Over a year ago

Ah sorry I just edited my post so I wouldn't get this answer :/

Ori Drori Over a year ago

See updated answer

Simon Berens Over a year ago

thanks! In the case where f has multiple such parameters, is there an extensible solution?

Simon Berens Over a year ago

links seem to be broken now

Ori Drori Over a year ago

See these answers -stackoverflow.com/a/67605309/5157454, stackoverflow.com/a/63029283/5157454

Simon Berens · Accepted Answer · 2022-03-12 00:43:54Z

0

I ended up making a simple wrapper function like so:

function ok<T extends string>(s: T): T {
  return s;
}

that I could then use like f(ok<S>('a')), because in my case there were multiple functions and types that I wanted to do this verification for, and making a separate wrapping function would be tedious and not maintainable.

edited Mar 12, 2022 at 0:43

answered Mar 12, 2022 at 0:38

Simon Berens

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2 Comments

Simon Berens Over a year ago

side note: Initially, intellij was giving me autocomplete suggestions for these strings (they were actually template literal types)--then I did something (I wish I knew what) and now it stopped :( Does anyone know how to ensure intellij is autocompleting them?

Simon Berens Over a year ago

it was because there were too many possibilities for my template literals

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