Creating a universal type for an arrow function in Typescript

I like to write my TypeScript functions in a functional style. When it comes to simple functions, I usually do something like this:

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

But when it comes to using generic types, things can get a bit more complicated. How can I maintain that simplicity with a function like this?

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

Just adding a generic type doesn't work as expected:

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

Is there a better way to achieve this?

Answer â„–1

Looking at your previous code snippet:

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

You specify that a is of type A, but you haven't assigned it to a particular generic type, which leads the compiler to default to using any.

To assign a specific generic type, you can do this:

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

Alternatively, you can define it like this:

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

If you want a to be more defined.

If you prefer a to remain generic, you'll have to declare the generic constraint on it as follows:

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

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