How do you create an async version of a TypeScript function type?

Question

I have a non-async function type in TypeScript and I want the same function type but async.

// this is a function type
type MyFunction = (x: number) => void;
// this is an async version of the same function type, created explicitly
type AsyncMyFunction = (x: number) => Promise<void>;

MyFunction could be any function, so more generically it's (...args: any[]) => any.

// how do I get the async version generically without knowing the actual function type?
type AsyncWrapper<T> = (/*args of T*/) => Promise</*return type of T*/>

If it works correctly, I should be able to use AsyncWrapper<MyFunction> in place of AsyncMyFunction because the AsyncWrapper will be able to copy the parameter types of T and wrap the return type of T in a Promise.

Is this possible?

Tobias S. · Accepted Answer · 2022-07-08 22:06:08Z

5

Here is a simple solution using the utility types Parameters and ReturnType.

type AsyncWrapper<T extends (...args: any) => any> = 
  (...args: Parameters<T>) => Promise<ReturnType<T>>

Playground

Side note: Nested Promises are always flattened in JavaScript. So nested Promise-Types also don't make much sense in TypeScript. If you may also pass functions that already return a Promise to this generic type, you may want to keep their return type as is.

type AsyncWrapper<T extends (...args: any) => any> = 
  (...args: Parameters<T>) => ReturnType<T> extends infer U extends Promise<any> 
    ? U 
    : Promise<ReturnType<T>>

Playground

edited Jul 8, 2022 at 22:06

answered Jul 8, 2022 at 19:42

Tobias S.

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1 Comment

Vala Over a year ago

While this works for simple functions I'm afraid it doesn't seem to work for functions with multiple declarations (overloaded functions). Not that it invalidates this answer, just worth noting for someone coming here later.

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