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无法用typescript定义泛型比较函数的类型

generics typescript

1

HMR · 技术社区 · 6 年前

我无法使此工作,无法确定如何将以下代码的类型解释为typescript:

interface hasLength {
  length: number;
}
type CompareFn = <T> (a: T, b: T) => boolean
const compare = (compareFn: CompareFn, a: hasLength, b: hasLength) => 
  compareFn(a, b)
const compareFn: CompareFn = (a, b) => a.length === b.length//error here
const otherCompare: CompareFn = (a, b) => a === b
console.log(compare(compareFn, [1], [2]))
console.log(compare(compareFn, 'a', 'b'))
console.log(compare(otherCompare, [1], [2]))

compare函数获取同一类型或接口的两个参数,并用于函数比较。错误是 Property 'length' does not exist on type 'T'.

1 回复 | 直到 6 年前

1

T.J. Crowder 6 年前

有两件事:

CompareFn 不是普通的 <T>
任何特定的 比较

下面是一个更正的版本,其中有注释,指出了更改/添加:

interface hasLength {
  length: number;
}

type CompareFn<T> = (a: T, b: T) => boolean
//            ^^^

const compare = (compareFn: CompareFn<hasLength>, a: hasLength, b: hasLength) => 
//                                   ^^^^^^^^^^^
  compareFn(a, b)
const compareFn: CompareFn<hasLength> = (a, b) => a.length === b.length
//                        ^^^^^^^^^^^
const otherCompare: CompareFn<any> = (a, b) => a === b
//                           ^^^^^
console.log(compare(compareFn, [1], [2]))
console.log(compare(compareFn, 'a', 'b'))
console.log(compare(otherCompare, [1], [2]))

On the playground

旁注:按照压倒性的惯例, hasLength 应该是 HasLength . 接口和类名以大写字母开头(同样,按惯例)。