Typescript / 类型安全柯里化函数

如何在打字稿中安全地输入柯里化函数？ 特别注意以下示例

interface Prop {
    <T, K extends keyof T>(name: K, object: T): T[K];
    <K>(name: K): <T>(object: T) => /* ?? */;
}

const prop: Prop = (key, object) => object[key];

const valid1 = prop('foo', { foo: 'hello' }); // string
const valid = prop('foo')({ foo: 'hello' });  // string

// `never`, since `baz` does not exist in { foo: string }
const invalid = prop('baz')({ foo: 'hello' }); // never

函数重载 https://www.typescriptlang.org/docs/handbook/functions.html#overloads

function prop<T, K extends keyof T>(name: K, obj: T): T[K]
function prop<K extends PropertyKey>(name: K):
    <T extends Record<K, unknown>>(obj: T) => T[K]
function prop(name: any, obj?: any) { 
    if (obj === undefined) {
        return (obj: any) => obj[name]
    } else {
        return obj[name]
    }
}
// weak types used in impl for simplicity, as they don't matter for the caller.
// also this function body is not really complex

const valid1 = prop('foo', { foo: 'hello1' }); // string
const valid2 = prop('foo')({ foo: 'hello2' });  // string
const invalid = prop('baz')({ foo: 'hello' }); // compile error, `baz` not in { foo: string } 

功能类型 https://www.typescriptlang.org/docs/handbook/interfaces.html#function-types

interface Prop {
    <T, K extends keyof T>(name: K, obj: T): T[K];
    <K extends PropertyKey>(name: K): <T extends Record<K, unknown>>(obj: T) => T[K]
}

const prop: Prop = (name: any, obj?: any) => {
    if (obj === undefined) {
        return (obj: any) => obj[name]
    } else {
        return obj[name]
    }
}
// weak types used here for simplicity like in first solution

const valid1 = prop('foo', { foo: 'hello1' }); // string
const valid2 = prop('foo')({ foo: 'hello2' });  // string
const invalid = prop('baz')({ foo: 'hello' }); // never
console.log(valid1, valid2) // hello1 hello2

Note:函数重载和函数类型不能完全互换使用（更多信息here https://github.com/microsoft/TypeScript/issues/33482 or here https://github.com/microsoft/TypeScript/issues/35338#issuecomment-558867975）。对于后者，可能需要用以下方式注释类型any在函数实现部分中考虑定义的调用签名中不兼容的返回类型 - 检查此举个例子。