Is it possible declare type optionally in typescript?-CodePudding

I'm looking for something like this...

type B = a ? 'apple' | 'grape' | 'orange' : 'apple' | 'grape'; // Of course, ERROR!

const x = (a: boolean, b: B) => console.log('foo')

How can I implement this? Is it possible in Typescript?

CodePudding user response：

You can indeed. To do this, you'll need to use generics and conditional types. Like this:

type Type1 = "apple" | "grape" | "orange";
type Type2 = "apple" | "grape";

const x = <A extends boolean>(a: A, b: A extends true ? Type1 : Type2) =>
  console.log("foo");

x(false, "orange"); // error
x(true, "orange"); // no error

CodePudding user response：

Please check this typescript document. https://www.typescriptlang.org/docs/handbook/2/conditional-types.html