如何：接受只读类型的泛型？

Question

给定列表上的一个动作

type DoSomethingWith<L extends any[]> = L

我想做的事情就是这样

const keys = {
  a: ['a', 'b', 'c'] as ['a', 'b', 'c'],
  d: ['d', 'e', 'f'] as ['d', 'e', 'f'],
}

type Keys = typeof keys

type KeysWithSomething = {
  [K in keyof Keys]: DoSomethingWith<Keys[K]>
}

但为了避免冗余（在可能更长的列表上）我希望能够这样写：

const keys = {
  a: ['a', 'b', 'c'] as const,
  d: ['d', 'e', 'f'] as const,
}

type Keys = typeof keys

type DoSomethingWith<L extends any[]> = L

type KeyKinds = {
  [K in keyof Keys]: DoSomethingWith<Keys[K]>
                                  // ^^^^^^^: Type '{ a: readonly ["a", "b", "c"]; d: readonly ["d", "e", "f"]; }[K]' does not satisfy the constraint 'any[]'.
}

错误是我尝试传递一个只读类型DoSomething，该类型期望通用列表类型（any[]）它们是指定DoSomething它也应该接受只读元素的方法吗？

Answer 1

readonly是的，您可以在通用约束中使用修饰符：

type DoSomethingWith<L extends readonly any[]> = L
//                             ^ add this

或者，在缩小范围readonly后，您可以采取相反的方式并删除标志：keysas const

type Mutable<T> = T extends object ? { -readonly [K in keyof T]: Mutable<T[K]> } : T

使用您的类型进行测试（Playground）：

type T1 = Mutable<Keys> // { a: ["a", "b", "c"]; d: ["d", "e", "f"]; }

type KeyKinds = {
    [K in keyof Keys]: DoSomethingWith<Mutable<Keys[K]>> // compiles now
}