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In my project I need a number of string literal types and arrays of allowed values for variuos tasks like typeguards.

This is what we have:

type Animal = 'cat' | 'dog' | 'rabbit' | 'snake'
const animals: Animal[] = ['cat', 'dog', 'rabbit', 'snake']

It works, but it requires to list keys several times in source code.

Another way is to use enums as suggested in Get array of string literal type values but it creates overhead in runtime as compiled code for enum is bigger then array.

I found another way to do it that have no runtime overhead, but I'm not sure if it will work in the future, as it may be a bug.

const notWidened = <T extends string>(val: T[]) => val

const animals = notWidened(['cat', 'dog', 'rabbit', 'snake'])
type Animal = typeof animals[0]

So the question is if it's safe to use this snippet or it is going to break in the future. Is there a better way to get both literal string type and array without duplication.

Vladimir Danchenkov
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  • [This question](https://stackoverflow.com/questions/44480644/typescript-string-union-to-string-array) may be related... see if my updated [answer](https://stackoverflow.com/a/45486495/2887218) helps you. – jcalz Sep 12 '17 at 12:26
  • But your method looks fine to me also. – jcalz Sep 12 '17 at 12:31
  • At least, I can replace `typeof animals[0]` with `typeof animals[number]`, which looks a lot better! – Vladimir Danchenkov Sep 12 '17 at 12:51
  • you know in 2.4 you can have string enums? https://www.typescriptlang.org/docs/handbook/release-notes/typescript-2-4.html – Ross Scott Sep 12 '17 at 13:16

2 Answers2

6

I can't find a good piece of documentation for this at the moment, but your notWidened() function works just fine and is not a bug. TypeScript will infer string literals for generic type variables if you constrain the type parameter to string or a subtype of string. So <T extends keyof U>, <T extends string>, <T extends 'a'|'b'>, etc. will infer string literals for T. (You can also infer number or boolean literals if you constrain T similarly).

So your code is just fine as far as I can see; the only thing I might do differently is

type Animal = (typeof animals)[number]

instead of 

type Animal = typeof animals[0]

since the 0th element of animals is actually 'cat', even though you've told TypeScript it is 'cat'|'dog'|.... Yours is fine, though.

As I commented above, if you want TypeScript to consider animals to be a tuple where animals[0] is of type 'cat', and animals[1] is of type 'dog', etc., you can use something like the function tuple() in tuple.ts (UPDATE July 2018, starting in TypeScript 3.0 the compiler will be able to infer tuple types automatically, so the function can be more succinct):

export type Lit = string | number | boolean | undefined | null | void | {};
export const tuple = <T extends Lit[]>(...args: T) => args;

const animals = tuple('cat', 'dog', 'rabbit', 'snake');
type Animal = (typeof animals)[number];  // union type

which might come in handy for you.

TL;DR: your code is fine.

Hope that helps; good luck!

jcalz
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3

Maybe you can use a const enum string, this has none runtime overhead. It compiles the values into the code (Playground):

const enum Animals {
  Cat = 'cat',
  Dog = 'dog',
  Rabbit = 'rabbit',
  Snake = 'snake'
}

alert(Animals.Cat)

// Generated JavaScript code: 
// alert("cat" /* Cat */);
Markus
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