如何在TypeScript中声明固定长度数组

ep6jt1vc  于 2022-12-19  发布在  TypeScript
关注(0)|答案(6)|浏览(506)

冒着证明我对TypeScript类型缺乏知识的风险--我有以下问题。
当你像这样声明一个数组的类型时...

position: Array<number>;

...它可以让你创建一个任意长度的数组,但是,如果你想要一个数组包含一个特定长度的数字,例如,x,y,z分量为3,你能为一个固定长度的数组创建一个类型吗?

position: Array<3>

任何帮助或澄清感谢!

6pp0gazn

6pp0gazn1#

JavaScript数组具有接受数组长度的构造函数:

let arr = new Array<number>(3);
console.log(arr); // [undefined × 3]

然而,这只是初始大小,没有限制更改:

arr.push(5);
console.log(arr); // [undefined × 3, 5]

TypeScript具有元组类型,允许您定义具有特定长度和类型的数组:

let arr: [number, number, number];

arr = [1, 2, 3]; // ok
arr = [1, 2]; // Type '[number, number]' is not assignable to type '[number, number, number]'
arr = [1, 2, "3"]; // Type '[number, number, string]' is not assignable to type '[number, number, number]'
z9gpfhce

z9gpfhce2#

元组方法:

此解决方案提供了基于元组的严格FixedLengthArray(又名SealedArray)类型签名。

    • 语法示例:**
// Array containing 3 strings
let foo : FixedLengthArray<[string, string, string]>

这是最安全的方法,考虑到它防止访问边界之外的索引

    • 执行情况:**
type ArrayLengthMutationKeys = 'splice' | 'push' | 'pop' | 'shift' | 'unshift' | number
type ArrayItems<T extends Array<any>> = T extends Array<infer TItems> ? TItems : never
type FixedLengthArray<T extends any[]> =
  Pick<T, Exclude<keyof T, ArrayLengthMutationKeys>>
  & { [Symbol.iterator]: () => IterableIterator< ArrayItems<T> > }
    • 测试:**
var myFixedLengthArray: FixedLengthArray< [string, string, string]>

// Array declaration tests
myFixedLengthArray = [ 'a', 'b', 'c' ]  // ✅ OK
myFixedLengthArray = [ 'a', 'b', 123 ]  // ✅ TYPE ERROR
myFixedLengthArray = [ 'a' ]            // ✅ LENGTH ERROR
myFixedLengthArray = [ 'a', 'b' ]       // ✅ LENGTH ERROR

// Index assignment tests 
myFixedLengthArray[1] = 'foo'           // ✅ OK
myFixedLengthArray[1000] = 'foo'        // ✅ INVALID INDEX ERROR

// Methods that mutate array length
myFixedLengthArray.push('foo')          // ✅ MISSING METHOD ERROR
myFixedLengthArray.pop()                // ✅ MISSING METHOD ERROR

// Direct length manipulation
myFixedLengthArray.length = 123         // ✅ READ-ONLY ERROR

// Destructuring
var [ a ] = myFixedLengthArray          // ✅ OK
var [ a, b ] = myFixedLengthArray       // ✅ OK
var [ a, b, c ] = myFixedLengthArray    // ✅ OK
var [ a, b, c, d ] = myFixedLengthArray // ✅ INVALID INDEX ERROR

(*)此解决方案需要启用noImplicitAny类型脚本configuration directive才能工作(通常推荐的做法)

阵列(ish)方法:

此解决方案的行为相当于Array类型的扩展,接受额外的第二个参数(数组长度)。不如 * 基于元组的解决方案 * 严格和安全。

    • 语法示例:**
let foo: FixedLengthArray<string, 3>

请记住,这种方法不会阻止您访问声明边界之外的索引并在其上设置值。

    • 执行情况:**
type ArrayLengthMutationKeys = 'splice' | 'push' | 'pop' | 'shift' |  'unshift'
type FixedLengthArray<T, L extends number, TObj = [T, ...Array<T>]> =
  Pick<TObj, Exclude<keyof TObj, ArrayLengthMutationKeys>>
  & {
    readonly length: L 
    [ I : number ] : T
    [Symbol.iterator]: () => IterableIterator<T>   
  }
    • 测试:**
var myFixedLengthArray: FixedLengthArray<string,3>

// Array declaration tests
myFixedLengthArray = [ 'a', 'b', 'c' ]  // ✅ OK
myFixedLengthArray = [ 'a', 'b', 123 ]  // ✅ TYPE ERROR
myFixedLengthArray = [ 'a' ]            // ✅ LENGTH ERROR
myFixedLengthArray = [ 'a', 'b' ]       // ✅ LENGTH ERROR

// Index assignment tests 
myFixedLengthArray[1] = 'foo'           // ✅ OK
myFixedLengthArray[1000] = 'foo'        // ❌ SHOULD FAIL

// Methods that mutate array length
myFixedLengthArray.push('foo')          // ✅ MISSING METHOD ERROR
myFixedLengthArray.pop()                // ✅ MISSING METHOD ERROR

// Direct length manipulation
myFixedLengthArray.length = 123         // ✅ READ-ONLY ERROR

// Destructuring
var [ a ] = myFixedLengthArray          // ✅ OK
var [ a, b ] = myFixedLengthArray       // ✅ OK
var [ a, b, c ] = myFixedLengthArray    // ✅ OK
var [ a, b, c, d ] = myFixedLengthArray // ❌ SHOULD FAIL
py49o6xq

py49o6xq3#

实际上,你可以用当前的 typescript 实现这一点:

type Grow<T, A extends Array<T>> = ((x: T, ...xs: A) => void) extends ((...a: infer X) => void) ? X : never;
type GrowToSize<T, A extends Array<T>, N extends number> = { 0: A, 1: GrowToSize<T, Grow<T, A>, N> }[A['length'] extends N ? 0 : 1];

export type FixedArray<T, N extends number> = GrowToSize<T, [], N>;

示例:

// OK
const fixedArr3: FixedArray<string, 3> = ['a', 'b', 'c'];

// Error:
// Type '[string, string, string]' is not assignable to type '[string, string]'.
//   Types of property 'length' are incompatible.
//     Type '3' is not assignable to type '2'.ts(2322)
const fixedArr2: FixedArray<string, 2> = ['a', 'b', 'c'];

// Error:
// Property '3' is missing in type '[string, string, string]' but required in type 
// '[string, string, string, string]'.ts(2741)
const fixedArr4: FixedArray<string, 4> = ['a', 'b', 'c'];
    • 编辑(很长时间后)**

这应该可以处理更大的大小(因为基本上它会按指数级增长数组,直到最接近2的幂):

type Shift<A extends Array<any>> = ((...args: A) => void) extends ((...args: [A[0], ...infer R]) => void) ? R : never;

type GrowExpRev<A extends Array<any>, N extends number, P extends Array<Array<any>>> = A['length'] extends N ? A : {
  0: GrowExpRev<[...A, ...P[0]], N, P>,
  1: GrowExpRev<A, N, Shift<P>>
}[[...A, ...P[0]][N] extends undefined ? 0 : 1];

type GrowExp<A extends Array<any>, N extends number, P extends Array<Array<any>>> = A['length'] extends N ? A : {
  0: GrowExp<[...A, ...A], N, [A, ...P]>,
  1: GrowExpRev<A, N, P>
}[[...A, ...A][N] extends undefined ? 0 : 1];

export type FixedSizeArray<T, N extends number> = N extends 0 ? [] : N extends 1 ? [T] : GrowExp<[T, T], N, [[T]]>;
j8yoct9x

j8yoct9x4#

现在说这个有点晚了,但如果您使用的是只读数组([] as const),这里有一种方法-

interface FixedLengthArray<L extends number, T> extends ArrayLike<T> {
  length: L
}

export const a: FixedLengthArray<2, string> = ['we', '432'] as const

const a值中添加或删除字符串会导致此错误-

Type 'readonly ["we", "432", "fd"]' is not assignable to type 'FixedLengthArray<2, string>'.
  Types of property 'length' are incompatible.
    Type '3' is not assignable to type '2'.ts(2322)

Type 'readonly ["we"]' is not assignable to type 'FixedLengthArray<2, string>'.
  Types of property 'length' are incompatible.
    Type '1' is not assignable to type '2'.ts(2322)

分别。
编辑(2022年5月13日):相关的未来TS功能-satisfies defined here

f5emj3cl

f5emj3cl5#

下面是一个基于TomaszGawel的answer的超短版本,其中包含了typescript v4.6

type Tuple<
  T,
  N extends number,
  R extends readonly T[] = [],
> = R['length'] extends N ? R : Tuple<T, N, readonly [T, ...R]>;

// usage
const x: Tuple<number,3> = [1,2,3];
x; // resolves as [number, number, number]
x[0]; // resolves as number

还有其他一些方法可以强制使用length属性的值,但效果并不好

// TLDR, don't do this
type Tuple<T, N> = { length: N } & readonly T[];
const x : Tuple<number,3> = [1,2,3]

x; // resolves as { length: 3 } | number[], which is kinda messy
x[0]; // resolves as number | undefined, which is incorrect
lstz6jyr

lstz6jyr6#

对于任何需要比@ThomasVo中正确处理非字面数字的解决方案更通用的解决方案的人:

type LengthArray<
        T,
        N extends number,
        R extends T[] = []
    > = number extends N
        ? T[]
        : R['length'] extends N
        ? R
        : LengthArray<T, N, [T, ...R]>;

我需要使用这个类型来正确地处理未知长度的数组。

type FixedLength = LengthArray<string, 3>; // [string, string, string]
type UnknownLength = LengthArray<string, number>; // string[] (instead of [])

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