如何组合交集和联合类型答案

【问题标题】：How to combine intersection and union types如何组合交集和联合类型
【发布时间】：2017-12-18 22:27:15
【问题描述】：

需要通过属性c“扩展”基本类型Base。以下code：

/* @flow */

export type A = 'a1' | 'a2';
export type B = | 'b1' | 'b2' | 'b3' | 'b4';


type Base = {
  type: A,
  a: number,
} | {
  type: B,
  b: number,
};

type Derived = {
  c: boolean;
} & Base; // #17

const f = (x: Derived) => { // #19
  if(x.type === 'a1') {
    x.a = 3; // #21
  }
  if(x.type === 'b1') {
    x.b = 3; // #24
  }
}

结果

19: const f = (x: Derived) => {
                  ^ intersection type. This type is incompatible with
17: } & Base;
        ^ union: object type(s)
21:     x.a = 3;
     ^ assignment of property `a`. Property cannot be assigned on any member of intersection type
21:     x.a = 3;
     ^ intersection
24:     x.b = 3;
     ^ assignment of property `b`. Property cannot be assigned on any member of intersection type
24:     x.b = 3;
     ^ intersection

除了将相同的道具c 添加到工会的两个成员之外，还有其他解决方案吗？谢谢！

【问题讨论】：

标签： ecmascript-6 flowtype

【解决方案1】：

您可以反转这一点，使 Derived 成为 BaseA 和 BaseB 的并集，并为两个基添加具有交集的公共属性 (working example)：

/* @flow */

export type A = 'a1' | 'a2';
export type B = | 'b1' | 'b2' | 'b3' | 'b4';

type Base = {
  c: boolean;
};

type BaseA = Base & {
  a: number,
  type: A,
};

type BaseB = Base & {
  b: number,
  type: B,
};

type Derived = BaseA | BaseB;

const f = (x: Derived) => {
  x.c = true;
  if(x.type === 'a1') {
    x.a = 3;
  }
  if(x.type === 'b1') {
    x.b = 3;
  }
}

【讨论】：