Idiomatic Typescript Enum Discriminated Union

Question

As of typescript 2.0 you can use discriminated unions with an enum as the discriminant like this:

export function getInstance(code: Enum.Type1, someParam: OtherType1): MyReturnType1;
export function getInstance(code: Enum.Type2, someParam: OtherType2): MyReturnType2;
export function getInstance(code: Enum, someParam: UnionOfOtherTypes): UnionOfReturnTypes {
  switch (code) {
    case Enum.Type1:
      return new ReturnType1(someParam as OtherType1);
    case Enum.Type2:
      return new ReturnType2(someParam as OtherType2);
  }
}

As of TypeScript 2.3

Is this the idiomatic way to do this?
Are we able to infer the type of someParam without casting?
Are we able to simplify the type definitions, maybe using generics, altering the function parameters, etc so we only need to define the final function?
Is it possible declare the functions as consts like: const getInstance = () => {};

score 0 · Accepted Answer · answered May 05 '17 at 05:24

0

Is this the idiomatic way to do this

No. if you need to use a type assertion e.g. someParam as OtherType1 it is unsafe.

More

Unsafety of type assertion : https://basarat.gitbooks.io/typescript/docs/types/type-assertion.html
The idiomatic way is to use discriminant properties instead of function overloading. Examples : https://basarat.gitbooks.io/typescript/docs/types/discriminated-unions.html

answered May 05 '17 at 05:24

basarat

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I read your book all the time :) thank you! Is it possible to infer the return type of some discriminated union based on a parameter? i.e. if code is type A i know the return type would be type B. – chris May 05 '17 at 05:28
Yes. All return statements for a union for the return type. e.g. if you return a `string` and a `number` in different sections of your code the return type is inferred to be `string | number` – basarat May 05 '17 at 05:35

Idiomatic Typescript Enum Discriminated Union

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