输入元组，以便功能链具有有效的参数和返回类型 [英] Type tuple so that function chain has valid parameter and return types

问题描述

我想为下面的这个元组/数组创建一个类型。

I'd like to create a type for this tuple / array below.

在有效的地方：

const funcs = [(a: string) => 1, (a: number) => 'A', (a: string) => 2]

这是无效的：

const funcs = [(a: string) => 1, (a: number) => 2, (a: string) => 3]

区别在于中间函数的返回类型从字符串更改为数字。

The difference being the return type for the middle function changed from string to number.

有可能吗？

type SubtractOne<T extends number> = [-1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62][T];
type AddOne<T extends number> = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62][T];

const funcs = [(a: string) => 1, (a: number) => 'A', (a: string) => 2]

type CheckFuncs<T extends any[]> = { [(K extends number) in keyof T]: T[AddOne<K>] }

type funcsType = typeof funcs
type funcsType2 = CheckFuncs<funcsType>

在研究期间，我遇到了一种使用映射进行索引的方法。是否可以使用它在K中添加或减去 AddOne ？那么我可以访问 T [K] 的 ReturnType 和 Parameter< T [k +1]> [0] ？

During my research I came across a way to index with a mapping. Is it possible to use this to add or subtract AddOne to K? then I can access the ReturnType of T[K] and the Parameter<T[k+1]>[0]?

l ayground

推荐答案

在给定索引类型的情况下，索引到元组的下一个或上一个值K 是我对元组休息/传播。考虑 Tail< T> ，它采用一个元组类型 T 并返回一个删除了第一个元素的新元组类型：

Indexing into the next or previous value of a tuple given an index type K is something I'd do with the support for tuple rest/spread. Consider Tail<T>, which takes a tuple type T and returns a new tuple type with the first element removed:

// Tail<[1,2,3]> is [2,3]
type Tail<T extends readonly any[]> =
    ((...a: T) => void) extends ((h: any, ...r: infer R) => void) ? R : never;

然后 T [K + 1] 可以表示为 Tail< T> [K] 。

这可能是您所提问题的答案，但我将继续介绍如何使用它：

That's probably the answer to your question as asked, but I'll continue a little to show how to use it:

这是我开始编写您的 CheckFuncs 类型的方式：

Here's how I might start writing your CheckFuncs type:

type CheckFuncs<T extends readonly ((x: any) => any)[]> = { [K in keyof T]:
    K extends keyof Tail<T> ? (
        [T[K], Tail<T>[K]] extends [(x: infer A) => infer R, (x: infer S) => any] ? (
            [R] extends [S] ? T[K] : (x: A) => S
        ) : never
    ) : T[K]
}

经过 T 并比较 T [ K] 与 Tail< T> [K] ，并转换 T [K] 变成一个可以正常工作的版本。然后这些（请注意，您需要 const 断言以保留元组类型）：

That walks through T and compares T[K] with Tail<T>[K], and converts T[K] into a version of itself that works. Then these (note that you need const assertions for the tuple type to be preserved):

const pass = [(a: string) => 1, (a: number) => 'A', (a: string) => 2] as const
const fail = [(a: string) => 1, (a: number) => 3, (a: string) => 2] as const

产生以下内容：

type passType = CheckFuncs<typeof pass>
// readonly [(a: string) => number, (a: number) => string, (a: string) => number]

type failType = CheckFuncs<typeof fail>
// readonly [(a: string) => number, (x: number) => string, (a: string) => number]

您可以创建一个与 CheckFuncs 像这样：

And you can make a function that works with CheckFuncs like this:

function useFuncs<T extends readonly ((x: any) => any)[]>(...t: CheckFuncs<T>) { }

useFuncs((a: string) => 1, (a: number) => 'A', (a: string) => 2); // okay
useFuncs((a: string) => 1, (a: number) => 3, (a: string) => 2); // error!
// ----------------------> ~~~~~~~~~~~~~~~~
// Argument of type '(a: number) => number' is not assignable to 
// parameter of type '(x: number) => string'.

当然，您想要编写或使用 CheckFuncs 可能有所不同；这只是一个例子。希望能有所帮助；祝你好运！

Of course the exact way you want to write or use CheckFuncs might be different; this is just an illustration. Hope that helps; good luck!

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