推断函数参数类型并将它们传播到Typescript中

Question

给定一个函数类型F，我想创建一个“在F之前可组合”的新类型，这意味着它接受并返回F所采用的相同参数。例如：

const stringToNumber = (s: string) => s.length;
const stringToString: ComposableOn<(s: string) => number> = (s: string) => s + s;

const composedResult = stringToNumber(stringToString('a'));

但是，我无法正确定义ComposableOn类型。以下是我尝试过的事情：

type ComposableBefore<F extends (...args: any) => any> = (args: Parameters<F>) => Parameters<F>;
type StoN = (s: string) => number;

const sToS: ComposableBefore<StoN> = (s: string) => s + s; // error: the type is [string] => [string] and not string => string

type ComposableBefore<F extends (...args: any) => any> = Parameters<F> extends Array<infer U>? (...args: Parameters<F>) => U: never;

const complex: ComposableBefore<(a: string, b: number, c: { d: number, e: string }) => number> = (a, b, c) => c; // not good either, since it can return a value of any type of the original function's argument types.

键入此内容的正确方法是什么？

Answer 1

您可以使用...将Parameters直接传播到函数签名中。至于返回，因为你想支持多个参数，返回类型应该是一个元组，因此你需要扩展函数的返回：

const stringToNumber = (s: string) => s.length;
const stringToString: ComposableBefore<(s: string) => number> = (s: string) => [s + s];

const composedResult = stringToNumber(...stringToString('a'));

type ComposableBefore<F extends (...args: any) => any> = (...args: Parameters<F>) => Parameters<F>;
type StoN = (s: string) => number;

您还可以考虑为单个参数函数添加一个特殊情况，如果这是常见用例：

const stringToNumber = (s: string) => s.length;
const stringToString: ComposableBefore<(s: string) => number> = (s: string) => s + s;

const composedResult = stringToNumber(stringToString('a'));

const multiStringToNumber = (s: string, s2: string) => s.length + s2.length;
const multiStringToString: ComposableBefore<typeof multiStringToNumber> = (s: string, s2: string) => [s + s, s2 + s2];

const multiComposedResult = multiStringToNumber(...multiStringToString('a', 'b'));

type ComposableBefore<F extends (...args: any) => any> =
    F extends (a: infer U) => any ? (...args: Parameters<F>) => U :
        (...args: Parameters<F>) => Parameters<F>;