Is there a way to create a new type from an union type with some it's options removed?

Question

Let us say we have

type U = A | B | C

and we need type U without some of it's options

function f<T option U>(u: U): U without T {...}

How can we express that

U has type union
T is an option of U
The returned type is like U but without some option

?

It is the concept of T being part of union U. It is not real ts syntax — Daniel San, Jul 04 '19 at 09:27
https://stackoverflow.com/questions/52475711/exclude-type-by-member-type — Roberto Zvjerković, Jul 04 '19 at 09:31
https://stackoverflow.com/questions/48750647/get-type-of-union-by-discriminant — Roberto Zvjerković, Jul 04 '19 at 09:36
Possible duplicate of [Get Type of Union By Discriminant](https://stackoverflow.com/questions/48750647/get-type-of-union-by-discriminant) — Mu-Tsun Tsai, Jul 04 '19 at 10:37

score 0 · Answer 1 · answered Jul 04 '19 at 10:43

Yes, use Exclude from stadard library.

type U = 'a' | 'b' | 'c';
type nonA = Exclude<U, 'a'>; // 'b' | 'c'

And for your function

function f<T extends U>(u: U): Exclude<U,T> {...}

https://www.typescriptlang.org/docs/handbook/release-notes/typescript-2-8.html#conditional-types

score 0 · Answer 2 · answered Jul 04 '19 at 11:09

0

100% working example

function withoutType<U, T extends U>(u: U) {
  return u as Exclude<U, T>;
}

type Union = string | number | Function;
let x = 5;
let y = withoutType<Union, Function>(x); // let x: string | number;

answered Jul 04 '19 at 11:09

Serg The Bright

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Is there a way to create a new type from an union type with some it's options removed?

2 Answers2