如何在Typescript中为箭头函数创建泛型类型

Question

I try to write typescript functions in the style that is most close to functional. For simple functions I can write:

type A = (value: number) => string;
const a: A = value => value.toString();

But what can I do with generic types? How can I type in that simple way following function?

function a<T>(value: T): T {
  return value;
}

If I try to simply add a generic type, it gives nothing:

type A = <T>(value: T) => T;
const a: A = value => value; // `value` implicitly has an `any` type

Is there any way to do it?

Answer 1

In your last snippet:

type A = <T>(value: T) => T;
const a: A = value => value;

You tell the compiler that a is of type A , but you don't bind it to a specific generic type which is why it uses any .

For example, you can set the generic type like so:

const a: A = (value: string) => value;

You can also do this:

type A<T> = (value: T) => T;
const a: A<string> = value => value;

If you want a to be specific.

If you want a to stay generic you'll need to declare the generic constraint on it as well:

const a: A = <T>(value: T) => value;