如何让TS推断通用构建器函数中的回调？

Question

考虑以下用例演示(playground)：

// A builder that can self-reference its keys using a ref function
declare function makeObj<K extends string>(
    builder: (ref: (k: K) => number) => Record<K, number>
): Record<K, number>;

// Not using `ref` for now. All good, K is inferred as <"x" | "y">.
const obj1 = makeObj(() => ({ x: 1, y: 2 }));

// Oops, now that we try to use `ref`, K is inferred as <string>.
const obj2 = makeObj(ref => ({ x: 1, y: ref("invalid key, only x or y") }));

// This works, but we'd want K to be automatically inferred.
const obj3 = makeObj<"x" | "y">(ref => ({ x: 1, y: ref("x") }));

那么，我应该如何编写makeObj来自动推断K呢？

Answer 1

尝试将Record<K, number>声明为第二个泛型类型。

playground

declare function makeObj<K extends string, T extends Record<K, number> = Record<K, number>>(
  builder: (ref: (k: K) => number) => T
): T

const obj1 = makeObj(() => ({ x: 1, y: 2 }));
const obj2 = makeObj(ref => ({ x: 1, y: ref("invalid key, only x or y") }));
const obj3 = makeObj(ref => ({ x: 1, y: ref("x") }));

限制

嗯，正如@kaya3在下面评论的那样。此解析只能推断返回类型。除非显式设置泛型类型，否则它仍然找不到无效的键。

// error will shown when given explicit generic type
const obj2 = makeObj<'x' | 'y'>(ref => ({x: 1, y: ref("invalid key, only x or y")}));

Answer 2

type Ref<Obj> = {
  ref<Key extends string & keyof Obj>
      (k:`${Key}`):Obj[Key]
};

declare function makeObject<Type>(obj: Type): Type & Ref<Type>;


const person = makeObject({
  firstName: "Saoirse",
  lastName: "Ronan",
  age: 26
});

person.ref('firstName') // (method) ref<"firstName">(k: "firstName"): string

// Argument of type '"firstNme"' is not assignable to 
// parameter of type '"firstName" | "lastName" | "age"'.ts(2345)
person.ref('firstNme') // ERROR

person.ref('age') // (method) ref<"age">(k: "age"): number

参考"String Unions in Types"

不幸的是，当您在makeObject调用中声明对象时，您不能访问ref。幸运的是，原生Javascript的近似性已经相当不错了。