Typescript 推断执行具有相同参数签名的可选函数参数的高阶函数的类型

Question

假设我有一个函数，它接受两个函数f 和g 作为参数并返回一个执行f 和g 的函数并返回一个带有结果的对象。我还想强制 f 和 g 具有相同的签名。这对于条件类型来说很容易：

type ArgumentTypes<F extends Function> = F extends (...args: infer A) => any ? A : never;
function functionPair<
    F extends (...args: any[]) => any,
    G extends (...args: ArgumentTypes<F>) => any
>
(f: F, g: G): (...args: ArgumentTypes<F>) => { f: ReturnType<F>, g: ReturnType<G> }
{
    return (...args: ArgumentTypes<F>) => ({ f: f(...args), g: g(...args) });
}

functionPair((foo: string) => foo, (bar: number) => bar); // Error, incompatible signatures, as expected
functionPair((foo: string) => foo, (bar: string) => bar.length); // (foo: string) => { f: string; g: number; }, as expected

现在，如果我想让f 和g 可选，并因此改变返回对象的形状怎么办？也就是说，如果 f 或 g 是 undefined，则结果对象中应该缺少它们的键：

functionPair(); // Should be () => {}
functionPair(undefined, undefined); // Should be () => {}
functionPair((foo: string) => foo); // Should be (foo: string) => { f: string }
functionPair(undefined, (bar: string) => foo.length); // Should be (bar: string) => { g: number }
functionPair((foo: string) => foo, (bar: string) => foo.length); // Should be (foo: string) => { f: string, g: number }, as before

我一直在尝试使用条件类型来实现这一点，但是在有条件地强制执行结果函数的形状时遇到了一些麻烦。到目前为止，这是我所拥有的（严格的空检查已关闭）：

function functionPair<
    A extends F extends undefined ? G extends undefined ? [] : ArgumentTypes<G> : ArgumentTypes<F>,
    F extends (...args: any[]) => any = undefined,
    G extends F extends undefined ? (...args: any[]) => any : (...args: ArgumentTypes<F>) => any = undefined
>
(f?: F, g?: G): (...args: A) =>
    F extends undefined
    ? G extends undefined ? {} : { g: ReturnType<G> }
    : G extends undefined ? { f: ReturnType<F> } : { f: ReturnType<F>, g: ReturnType<G> }
{ /* implementation... */ }

const a = functionPair(); // () => {}, as expected
const b = functionPair((foo: string) => foo); // (foo: string) => { f: string; }, as expected
const c = functionPair((foo: string) => foo, (bar: number) => bar); // Error, incompatible signatures, as expected
const d = functionPair((foo: string) => foo, (bar: string) => bar.length); // (foo: string) => { f: string; g: number; }, as expected

const e = functionPair(undefined, undefined); // INCORRECT! Expected () => {}, got (...args: unknown[] | []) => {} | { f: any; } | { g: any; } | { f: any; g: any; }
const f = functionPair(undefined, (bar: string) => bar.length); // INCORRECT! Expected (bar: string) => { g: number; } but got (...args: unknown[] | [string]) => { g: number; } | { f: any; g: number; }

---

顺便说一句，我知道这在技术上可以通过重载实现，如下所示，但我真的很想了解如何在没有它们的情况下做到这一点。

function functionPairOverloaded(): () => {}
function functionPairOverloaded(f: undefined, g: undefined): () => {}
function functionPairOverloaded<F extends (...args: any[]) => any>(f: F): (...args: ArgumentTypes<F>) => { f: ReturnType<F> }
function functionPairOverloaded<G extends (...args: any[]) => any>(f: undefined, g: G): (...args: ArgumentTypes<G>) => { g: ReturnType<G> }
function functionPairOverloaded<F extends (...args: any[]) => any, G extends (...args: ArgumentTypes<F>) => any>(f: F, g: G): (...args: ArgumentTypes<F>) => { f: ReturnType<F>, g: ReturnType<G> }
function functionPairOverloaded<F extends (...args: any[]) => any, G extends (...args: any[]) => any>(f?: F, g?: G) { /* implementation... */ }

Answer 1

假设你打开了--strictNullChecks，我想我会这样做：

type Fun = (...args: any[]) => any;
type FunFrom<F, G> = F extends Fun ? F : G extends Fun ? G : () => {};
type IfFun<F, T> = F extends Fun ? T : never;
type Ret<T> = T extends (...args: any[]) => infer R ? R : never

declare function functionPair<
  F extends Fun | undefined = undefined,
  G extends ((...args: (F extends Fun ? Parameters<F> : any[])) => any) 
    | undefined = undefined
>(
  f?: F, 
  g?: G
): (...args: Parameters<FunFrom<F, G>>) => {
  [K in IfFun<F, 'f'> | IfFun<G, 'g'>]: K extends 'f' ? Ret<F> : Ret<G> 
};

这相当难看，但它确实为您提供了您正在寻找的行为：

const a = functionPair(); // () => {}, as expected
const b = functionPair((foo: string) => foo); // (foo: string) => { f: string; }, as expected
const c = functionPair((foo: string) => foo, (bar: number) => bar); // Error, incompatible signatures, as expected
const d = functionPair((foo: string) => foo, (bar: string) => bar.length); // (foo: string) => { f: string; g: number; }, as expected
const e = functionPair(undefined, undefined); // () => {}, as expected
const f = functionPair(undefined, (bar: string) => bar.length); // (bar: string) => { g: number; }, as expected

我决定只使用两个类型参数F 和G，而不是A 使用Parameters<FunFrom<F, G>>。请注意，Parameters 是一个内置类型函数，类似于您的ArgumentTypes。

另外，对于返回函数的返回类型，我做了一个有点难看的映射类型。我最初计划做类似IfFun<F, {f: Ret<F>}> & IfFun<G, {g: Ret<G>}> 的事情，这（我相信）更容易理解，但生成的类型{f: X, g: Y} 比交集{f: X} & {g: Y} 更好。

无论如何，希望对您有所帮助。祝你好运！

---

如果您希望能够关闭 --strictNullChecks，那么定义会变得更加复杂：

type Fun = (...args: any[]) => any;
type AsFun<F> = [F] extends [Fun] ? F : never
type FunFrom<F, G> = AsFun<IfFun<F, F, IfFun<G, G, () => {}>>>;
type IfFun<F, Y, N=never> = F extends undefined ? N : 
  0 extends (1 & F) ? N : F extends Fun ? Y : N;
type Ret<T> = T extends (...args: any[]) => infer R ? R : never

declare function functionPair<
  F extends Fun | undefined = undefined,
  G extends ((...args: IfFun<F, Parameters<F>, any[]>) => any)
  | undefined = undefined
  >(
    f?: F,
    g?: G
  ): (...args: Parameters<FunFrom<F, G>>) => {
    [K in IfFun<F, 'f'> | IfFun<G, 'g'>]: K extends 'f' ? Ret<F> : Ret<G>
  };

不同之处在于IfFun<>需要能够区分功能与undefined和any，当你关闭--strictNullChecks时，这两者都会在不幸的地方弹出。这是因为undefined extends Function ? true : false 开始返回true，并且当您将手动undefined 值传递给函数时，any 开始被推断。区分undefined 相当简单，因为Function extends undefined ? true : false 仍然是false，但区分any 很烦人，并且涉及到一些funny business。

再次祝你好运！