是否可以在 TypeScript 中的元组上使用 Array.prototype.map()，同时保留返回类型中的元组长度？

我希望如果我使用内置的map在 TypeScript 中对长度为 N 的元组执行函数，则返回类型也将是长度为 N 的元组（元素的类型可能不同，具体取决于传递给的函数）map）。相反，返回类型只是回调函数返回的任何类型的标准可变长度数组。元组的长度丢失。我编写了一个自定义函数来执行我想要的操作，但我想知道是否有更好的方法来逃避我。我正在努力提高对 TypeScript 的理解。我在代码下方添加了一个 TypeScript Playground 链接（具有相同的代码）。感谢您的帮助！

const nums = [1, 2, 3] as const;

// result1 type is string[]
const result1 = nums.map((num) => num.toString());

// so this won't compile
const result2: [string, string, string] = nums.map((num) => num.toString());

// a type assertion would work, but I'd rather not use one...
const result3 = nums.map((num) => num.toString()) as [string, string, string];

// ...because this also compiles yet the type of result4 doesn't match its value
const result4 = nums.map((num) => num.toString()) as [string, boolean, number, symbol, string, number];

// result5 type is [string, string, string]
const result5 = map(nums, (num) => num.toString());

// custom map function that yields the correct return type when used on a tuple
function map<A extends readonly [...any[]], B>(values: A, func: (value: A[number]) => B): { [K in keyof A]: B } {
    return values.map(func) as unknown as { [K in keyof A]: B };
}

See on: TypeScript 游乐场 https://www.typescriptlang.org/play?#code/MYewdgzgLgBGCuBbCMC8MDaBGANDATHgMwC6MAhiqJFANwBQ9A9EzAE4CmE8ANlFjCgBPAA4cYASxTQ2EsAHMMJetWjsuvfmjhIIAOkTkRACmMJEASjQA+HYj1QQAZSiyFxixYbNWEEIIALKRgAd3AAclhQRBEJHg4VcDVObj58AC5MGTl5PGyFPNccsnRzfUMTMyQrVFtzB2ci909vFgpBUXFKCA42KAlwUJBeABMhtgBrPAAjeFgASXCxtnIoAN64EFh4HphwDj1DxJp1VKgibTKDI1NzGrqkBpc3eQ8rSiym3Jh879+SVqsQ56aYcYDkHbiNbBcg8PwwaKxeIoIQcWBrKGdPYAM1OmgALDARiAuGBIjBDFBgAFJFAUAA3WHwBKqWApAmXXTXSp3Gx2J5fN4UFAYX4zEAgeLkMB4cygth5ISIaaSwovWVIeUAxhtdl8ACsHTEkhFYp+XzVxWOyQ0Bu0FSqyDwjvu-Mczxyb0BCJ2jkQFKMMGx8DAwH6gzWqxgQgkHB4IxQGIRIDYnDD6ig8DYYCN4hC6xzkLGg3IgngIni9GDofDOYqAB4AIIwDgADygHDACfU5GJYB4QkwwOlQiUJDwACFrMZGTxmRBMo28NXgJkZ0yOIuMHLeiRXROLJkAN6YADSkhzEw4QhAuMbJEyE5gAF8YEf6DBPxmsznZ-PucYK7vCgIYTGAIAhDmHwnhg55yDAV43neD4wE+z4MM+9BAA

请尝试下一步：

type A = readonly [1, 2, 3]

const nums: A = [1, 2, 3];
const toStr = <T extends number>(num: T) => num.toString() as `${T}`

const result2 = nums.map(toStr); // ("3" | "1" | "2")[]

我相信这不是最好的解决方案，因为你仍然有 ('3'|'2'|'1')[] 而不是 ['1','2','3']，但这是某种向前一步

我很高兴在这里看到其他解决方案。很有趣的问题）

仅适用于 T.S 4.1

UPDATE

在以下的帮助下可以创建您想要的类型超载 https://www.typescriptlang.org/docs/handbook/2/functions.html#function-overloads and 可变元组类型 https://www.typescriptlang.org/docs/handbook/release-notes/typescript-4-0.html#variadic-tuple-types

type Json =
    | string
    | number
    | boolean
    | { [prop: string]: Json }
    | Array<Json>

type CustomMap<T extends ReadonlyArray<Json>> = {
    [Index in keyof T]: { elem: T[Index] }
}

function map<Elem extends Json, List extends Elem[], Return>(list: [...List], callback: (value: Elem) => Return): CustomMap<List>
function map<Elem extends Json>(list: Elem[], callback: (elem: Elem) => unknown) {
    return list.map(callback)
}

// const result2: [{
//     elem: 1;
// }, {
//     elem: 2;
// }, {
//     elem: 3;
// }]
const result2 = map([1, 2, 3], elem => ({ elem }));

操场 https://tsplay.dev/mqQDkm