javascript - typescript /类型安全的 curry 函数

Question

如何在 typescript 中安全地键入 curry 函数？
特别关于以下示例

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

Answer 1

function overload

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 } 

Sample

function type

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

Sample
注意:函数重载和函数类型不能完全互换使用(更多信息 here 或 here)。对于后者，可能需要使用 any 注释类型在函数实现部分解释定义的调用签名中不兼容的返回类型 - 检查此 playground例如。