Typescript Currying and Async Functions

Question

In typescript writing definitions for manually curried functions. They work when the final return value is a non-promise, but fail when it is a promise. For example, this works:

function test (
  a: number,
): (
  b: number,
) => number
function test (
  a: number,
  b: number,
 ):  number
function test (a, b?) {
  if (b === undefined)
    return function (b: number) { test(a, b) }
  return a + b
}

While this fails:

function asynctest (
  a: number,
): (
  b: number,
) => Promise<number>
function asynctest (
  a: number,
  b: number,
 ):  Promise<number>
async function asynctest (a, b?) {
  if (b === undefined)
    return function (b: number) { return asynctest(a, b) }
  return await a + b
}

VSCode's built in type inspector suggests that the type returned by asynctest when provided 1 argument is (b: number) => Promise<number> as expected.

Tim Hope · Accepted Answer · 2018-01-27 04:51:14Z

During the writing of this question I solved it.

An async function must return a promise, so in the case of one argument being passed, rather than returning a function, it is returning a Promise of a function. So the correct async-await version is:

function asynctest (
  a: number,
): Promise<(
  b: number,
) => Promise<number>>
function asynctest (
  a: number,
  b: number,
 ):  Promise<number>
async function asynctest (a, b?) {
  if (arguments.length === 1)
    return function (b: number) { return asynctest(a, b) }
  return await a + b
}

The alternative version is using promises directly rather than using async-await (I think this is neater):

function promisetest (
  a: number,
): (
  b: number,
) => Promise<number>
function promisetest (
  a: number,
  b: number,
 ):  Promise<number>
function promisetest (a, b?) {
  if (arguments.length === 1)
    return function (b: number) { return promisetest(a, b) }
  return new Promise(a + b)
}

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