函数参数中的TypeScript Array Union Type

Question

我有一个联合类型的各种特定长度的数组:

[ number ] | [ number, number ] | [ number, number, number, number ]

如您所见，对于一个元素、两个元素或四个元素的数组都有要求。

我正在尝试创建一个包含具有这些长度之一的函数的对象。我该如何编写类型定义以实现这一点？

TS playground

例子:

const people: {
    name: string,
    address: Address,
    work: (numbers: [ number ] | [ number, number ] | [ number, number, number, number ]) => any
}[] = [
    {
        name: "Bob",
        address: new Address(),
        work: function(numbers: [ number ]): number {
            // Implementation returning number
        }
    },
    {
        name: "Ashley",
        address: new Address(),
        work: function(numbers: [ number, number, number, number ]): boolean {
            // Implementation returning boolean
        }
    },
    {
        name: "Michael",
        address: new Address(),
        work: function(numbers: [ number, number ]): number {
            // Implementation returning number
        }
    },
]

目前，它给我错误:

类型 '(numbers: [number]) => number' 不可分配给类型 '(numbers: [number] | [number, number] | [number, number, number, number]) => any'。参数类型“numbers”和“numbers”不兼容。输入'[数字] | [数字，数字] | [number, number, number, number]' 不可分配给类型 '[number]'。类型“[number, number]”不可分配给类型“[number]”。源有 2 个元素，但目标只允许 1.ts(2322)

-------- 编辑--------

我应用了评论中的建议，并将所有可能的调用都变成了单独的函数联合，而不是数组联合:

const people: {
    name: string,
    address: Address,
    work: ((numbers: [ number ]) => any) | ((numbers: [ number, number ]) => any) | ((numbers: [ number, number, number, number ]) => any)
}[] = [

现在尝试从此数组调用函数时:

 people[1].work([2, 8, 6, 4])

它现在抛出以下错误:

在 VSCode 中我发现这是为什么:

“'[number] & [number, number] & [number, number, number, number]' 的交集被缩减为 'never'，因为属性 'length' 在某些成分中具有冲突类型。”

Answer 1

已更新你需要在这里使用双方差

class Address { }

type Tuple<
  N extends number,
  Item = number,
  Result extends Array<unknown> = [],
  > =
  (Result['length'] extends N
    ? Result
    : Tuple<N, Item, [...Result, number]>
  )


interface WorkFn {
  work(numbers: Tuple<1> | Tuple<2> | Tuple<4>): any
}

interface Person extends WorkFn {
  name: string,
  address: Address,
}

const people: Person[] = [
  {
    name: "Bob",
    address: new Address(),
    work(numbers: Tuple<1>) {
      const [myNumber] = numbers;

      return myNumber * 6
    }
  },
  {
    name: "Ashley",
    address: new Address(),
    work: function (numbers: Tuple<4>): boolean {
      const [myNumber, anotherNumber, someNumber, replaceNumber] = numbers;

      return myNumber === anotherNumber && someNumber === replaceNumber;
    }
  },
  {
    name: "Michael",
    address: new Address(),
    work: function (numbers: Tuple<2>): number {
      const [myNumber, anotherNumber] = numbers;

      return myNumber * anotherNumber;
    }
  },
]

TypeScript playground

Here您可以找到方法类型和箭头函数类型之间的区别以及关于 bivariance

另外，请注意它不是 100% 安全的