TypeScript：是否可以定义一个接受不同类型 arguments 的可变参数 function？

Question

I want to write a generic function that accepts variable number of arguments that may have different types and returns a tuple based on those arguments.

Here is an example in JavaScript:

function evaluate (...fns) {
  return fns.map(fn => fn())
}

evaluate(
  () => 10
) // [ 10 ]

evaluate(
  () => 10,
  () => 'f',
  () => null
) // [ 10, 'f', null ]

And in TypeScript I need to somehow convert the spread argument tuple to a resulting one:

function evaluate<T1, T2 ... Tn> (
  ...fns: [() => T1, () => T2 ... () => Tn]
): [T1, T2 ... Tn] {
  return fns.map(fn => fn()) as [T1, T2 ... Tn]
}

evaluate(
  () => 10
) // [ 10 ]: [number]

evaluate(
  () => 10,
  () => 'f',
  () => null
) // [ 10, 'f', null ]: [number, string, null]

I've tried a naive approach of creating an overload for all reasonable lengths of tuple:

function evaluate<T1> (
  fn1: () => T1
): [T1]
function evaluate<T1, T2> (
  fn1: () => T1,
  fn2: () => T2
): [T1, T2]
function evaluate<T1, T2, T3> (
  fn1: () => T1,
  fn2: () => T2,
  fn3: () => T3
): [T1, T2, T3]
function evaluate<T1, T2, T3> (
  ...fns: Array<(() => T1) | (() => T2) | (() => T3)>
): [T1] | [T1, T2] | [T1, T2, T3] {
  return fns.map(fn => fn()) as [T1] | [T1, T2] | [T1, T2, T3]
}

But it looks horribly, doesn't scale well and causes issues with a more complex function body.

Is there any way this could be done dynamically? Thanks!

Answer 1

The easiest way to implement this is to make evaluate() generic in its arraylike output type T (intended to be a tuple type ), and then represent the fns rest parameter as a mapped type on T , noting that mapped array/tuple types are also array/tuple types :

function evaluate<T extends any[]>(
  ...fns: { [I in keyof T]: () => T[I] }
) {
  return fns.map(fn => fn()) as T;
}

Note that the type assertion as T is necessary because the compiler cannot see that fns.map(fn => fn()) will have the effect of converting an array/tuple of function types to the array/tuple of corresponding return types. See Mapping tuple-typed value to different tuple-typed value without casts for more information.

Because {[I in keyof T]: () => T[I]} is a homomorphic mapped type where we are mapping directly over keyof T (see What does "homomorphic mapped type" mean? for more information), the compiler is able to infer T from it (linked page is deprecated, but still accurate and no new page exists ♂️).

Let's see it in action:

const x = evaluate(() => 10);
// const x: [number]

const y = evaluate(
  () => 10,
  () => 'f',
  () => null
)
// const y: [number, string, null]

Looks good. The compiler sees that x is of type [number] and y is of type [number, string, null] . It also behaves reasonably in cases where you pass in a rest argument of unknown order/length:

const fs = [() => "a", () => 3];
// const fs: ((() => string) | (() => number))[]

const z = evaluate(...fs);
// const z: (string | number)[]

Here fs is of the type Array<(()=>string) | (()=>number)> Array<(()=>string) | (()=>number)> , and so z is of the analogous type Array<string | number> Array<string | number> .

Playground link to code