使用 TypeScript，我可以输入 getProperty 的柯里化版本吗

示例来自https://www.typescriptlang.org/docs/handbook/advanced-types.html

function getProperty<T, K extends keyof T>(o: T, name: K): T[K] {
    return o[name]; // o[name] is of type T[K]
}

柯里化版本：

function curriedGetProperty<T, K extends keyof T>(name: K): (o: T) => T[K] {
    return (o: T) => o[name]; // o[name] is of type T[K]
}

const record = { id: 4, label: 'hello' }

const getId = curriedGetProperty('id') // Argument of type '"id"' is not assignable to parameter of type 'never'.

const id = getId(record)

编辑 TypeScript >= 4.1.5

const makeGetter = <TKey extends string>(key: TKey) => <TObject extends { [P in TKey]?: unknown }>(object: TKey extends keyof TObject ? TObject : `${TKey} is missing as property of object`) => (object as TObject)[key];

const getId = makeGetter('id');

const a: unknown = getId({})
const b: number = getId({id: 1})
const c: number | undefined = getId({} as { id?: number})

编译器会抱怨getId({})并附有有用的错误消息。

使用 TypeScript3.0.3我能够做到这一点：

function composeGetter<K extends string>(prop: K) {
    function getter<T extends { [P in K]?: any }>(object: T): T[typeof prop]
    function getter<T extends { [P in K]: any }>(object: T) {
        return object[prop]
    }

    return getter
}