typescript - 仅返回提供的键的部分类型的泛型函数的返回类型是什么？

Question

所以我的目标是有这样的东西，这样返回类型只包括传递给函数的键:

function doTheThings<T, U = Partial<T>>(values: U): { [P in keyof U]: U[P] };

目前，我得到的假设是我的参数和返回类型都是独立的 Partial<T> ，因此参数上不存在的键在返回类型上可用，如下所示:

type Thing = { name: string, age: number };
const seuss = doTheThings<Thing>({ name: 'seuss'});

// Should work, but TypeScript raises error
//   Type 'string | undefined' is not assignable to type 'string'.
//   Type 'undefined' is not assignable to type 'string'.
const name: string = seuss.name;

// Should work, but TypeScript raises error
//   Type 'number | undefined' is not assignable to type 'undefined'.
//   Type 'number' is not assignable to type 'undefined'.
const age: undefined = seuss.age;

我今天在 TypeScript 中尝试做的事情是否可行？
见操场:
https://www.typescriptlang.org/play?#code/GYVwdgxgLglg9mABAEzgFQBYFNMzAcwGcAeNAGkQFVEBeRABQEMAnWRgG1ID4uAKANw4gshAFxUAlOIDeiANr1EeRAGssATzjAqAXXGUFOxAF9E0gFCJEzLFBDMkg9sMIBuc8fPmo6gA5ZEXAJaM0QwRgBbLHFCKGY8fApGfGiwkAiAIyxmE3dzCARYxEIsEEJCENRMHAwEkiD8PllwqPEAchKywjbjCTyAen7EKFqKp2FijDgQdmRELMRGYriE4rwIAJgoRAB3Rgr152QsOf3FpBZ8dKwwbag4YexEUEhYBHzC7ZbU2Pjguk65QAdN8BkMRjAxkIAoQpjM5gtwMdgHgTmtIJttnsKmA4NtDiBjqcKowLswrlFbsMHiMAi9oPAwB8wEVkqkkVgUWA0QDSsC2e4gA

Answer 1

问题是推理如何与显式和默认类型参数一起工作。如果您指定一个类型参数，则不会对参数的其余部分进行推理(将使用它们的默认值)
通常的解决方法是使用函数柯里化(Currying):

function doTheThings<T>() {
  return function <U extends  Partial<T>>(values: U): U {
    return values;
  }
}

type Thing = { name: string, age: number };

const seuss = doTheThings<Thing>()({ name: 'seuss'});

// this value should be a string since it was included as an argument to the function
const name: string = seuss.name;

//  error age does not exist 
const age: undefined = seuss.age;

Playground Link
上面的函数只会返回一个带有传入键的对象。如果你想保留未传递的键，你可以与 Partial<T> 相交。 :

function doTheThings<T>() {
  return function <U extends  Partial<T>>(values: U): U & Partial<T> {
    return values;
  }
}

const age: number | undefined = seuss.age;

Playground Link