TypeScript 面试题（山月）

在 Typescript 中如何实现类型标记 Pick 与 Omit

原文：https://q.shanyue.tech/fe/ts/695.html

更多描述

有以下测试用例

interface User {
  id: number;
  age: number;
  name: string;
}

// 相当于: type PickUser = { age: number; name: string; }
type OmitUser = Omit<User, "id">;

// 相当于: type PickUser = { id: number; age: number; }
type PickUser = Pick<User, "id" | "age">;

Issue

欢迎在 Gtihub Issue 中回答此问题: Issue 695(opens new window)

Author

回答者: shfshanyue(opens new window)

type Pick<T, K extends keyof T> = {
  [P in K]: T[P];
};

type Exclude<T, U> = T extends U ? never : T;

type Omit<T, K extends keyof any> = Pick<T, Exclude<keyof T, K>>;

interface User {
  id: number;
  age: number;
  name: string;
}

// 相当于: type PickUser = { age: number; name: string; }
type OmitUser = Omit<User, "id">;

// 相当于: type PickUser = { id: number; age: number; }
type PickUser = Pick<User, "id" | "age">;

Author

回答者: Asarua(opens new window)

type MyPick<O extends object, K extends keyof O> = {
  [P in K]: O[P];
};

type MyOmit<O extends object, K extends keyof O> = MyPick<
  O,
  Exclude<keyof O, K>
>;

type MyOmit2<O extends object, K extends keyof O> = {
  [P in Exclude<keyof O, K>]: O[P];
};

什么是协变与逆变

原文：https://q.shanyue.tech/fe/ts/713.html

Issue

欢迎在 Gtihub Issue 中回答此问题: Issue 713(opens new window)

Author

回答者: shfshanyue(opens new window)

协变与逆变(Covariance and contravariance )是在计算机科学中，描述具有父/子型别关系的多个型别通过型别构造器、构造出的多个复杂型别之间是否有父/子型别关系的用语。

Author

回答者: Carrie999(opens new window)

https://github.com/sl1673495/blogs/issues/54

Author

回答者: Carrie999(opens new window)

https://www.zhihu.com/question/38861374

在 ts 中如何实现 Partial

原文：https://q.shanyue.tech/fe/ts/714.html

更多描述

实现 Partial，使得 Object 所有的属性变为可选属性。

PS: Partial 已经在 TS 中原生实现，见文档: https://www.typescriptlang.org/docs/handbook/utility-types.html#partialtype(opens new window)

type User = {
  id: number;
  age: number;
  name: string;
};

// Output:
// type PartialUser = {
//   id?: number | undefined;
//   age?: number | undefined;
//   name?: string | undefined;
// }
type PartialUser = Partial<User>;

以下是使用案例

interface Todo {
  title: string;
  description: string;
}

function updateTodo(todo: Todo, fieldsToUpdate: Partial<Todo>) {
  return { ...todo, ...fieldsToUpdate };
}

const todo1 = {
  title: "organize desk",
  description: "clear clutter",
};

const todo2 = updateTodo(todo1, {
  description: "throw out trash",
});

Issue

欢迎在 Gtihub Issue 中回答此问题: Issue 714(opens new window)

Author

回答者: shfshanyue(opens new window)

type Partial<T> = {
  [P in keyof T]?: T[P];
};

在 ts 中什么是 infer，并实现 Parameters 与 ReturnType

原文：https://q.shanyue.tech/fe/ts/715.html

Issue

欢迎在 Gtihub Issue 中回答此问题: Issue 715(opens new window)

Author

回答者: heretic-G(opens new window)

type Parameters<T extends (...args: any[]) => unknown> = T extends (
  ...args: infer R
) => unknown
  ? R
  : never;

Author

回答者: zhimazz(opens new window)

Parameters 是啥

Author

回答者: Asarua(opens new window)

Parameters 是啥

取得某个函数的参数类型的高级类型

Author

回答者: iceycc(opens new window)

Parameters 的作用是用于获得函数的参数类型组成的元组类型。

type Parameters<T extends (...args: any) => any> = T extends (
  ...args: infer P
) => any
  ? P
  : never;

type A = Parameters<() => void>; // []
type B = Parameters<typeof Array.isArray>; // [any]
type C = Parameters<typeof parseInt>; // [string, (number | undefined)?]
type D = Parameters<typeof Math.max>; // number[]