如何在打字稿中编写“反转"类型以反转元组的顺序

问题描述

type a = [1,2,3]
type Invert<T extends any[] & {'0': any}> = ???
type b = Invert<a> // should yield [3,2,1]

我一直想弄清楚元组的 Invert 类型的定义，也是 Init 和 Last 类型，尽管它们可以相互构造

I am stucked to figure out the definition of Invert type of a tuple,also an Init and Last type, although they may be constructed of each others

我尝试过的:

1. 在函数参数定义中定位类型并推断Rest部分，这种方法只得到带有rest参数的Tail部分

1. position the type in a function param definition and infer the Rest part, this approach only got the Tail part with rest params

推荐答案

从 typescript 4.0 开始处理 Reverse 类型的代码比以前容易得多，说

as of typescript 4.0code to approach a Reverse type is much easier than before, saying

type Reverse<T extends any[], R extends any[] = []> =  ReturnType<T extends [infer F, ...infer L] ? () => Reverse<L,[F,...R]> : () => R>

一些解释:

1. 类型别名Reverse中的直接引用类型Reverse会导致循环引用错误.
2. wrap Reverse type in function type (()=> Reverse) 将选择退出循环引用错误
3. 如果类型表达式可以静态解析，编译器会尝试解析它，所以 ReturnTypeTup<L,[F,...R]>> 不会成功，但是ReturnType 会做

1. direct reference type Reverse in typealias Reverse will result in circularly references error.
2. wrap Reverse type in function type (()=> Reverse) will opt-out the circularly references error
3. if type expression could resolve staticly, the compiler will try to resolve it, so ReturnType<() => Tup<L,[F,...R]>> wont do the trick, butReturnType<ConditionalType> will do

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