TypeScript:从可区分的联合中推断类型以创建 map ？

Question

enum ShapeType { Circle, Square }

interface Circle {
    kind: ShapeType.Circle,
    radius: number
}

interface Square {
    kind: ShapeType.Square,
    sideLength: number
}

type Shape = Circle | Square

type WhatShape<T extends ShapeType>   // <-- can we do without WhatShape?
    = T extends ShapeType.Circle ? Circle
    : T extends ShapeType.Square ? Square
    : never;

type ShapeMap = {
    [K in ShapeType]: (s: WhatShape<K>) => void
}


const handlers: ShapeMap = {
    [ShapeType.Circle]: (c: Circle) => {
        console.log(c.radius)
    },
    [ShapeType.Square]: s/*inferred!*/ => {
        console.log(s.sideLength)
    },
}

playground

这可行，但必须维护 WhatShape 以将枚举映射回形状类型，这有点烦人。

有没有什么方法可以让 TS 从 kind 推断出我们的 ShapeMap 的参数类型？

Answer 1

您可以使用 the Extract<T, U> utility type查找工会成员 T可分配给类型 U .在你的情况下，T将是 discriminated union输入 Shape , 和 U将是一个对象类型，其 kind属性是 ShapeType 的特定成员你正在尝试映射。所以WhatShape<T>可以定义为:

type WhatShape<T extends ShapeType> = Extract<Shape, { kind: T }>

它的行为与您现有的版本相同，无需与 Shape 分开维护:

type ShapeMap = {
    [K in ShapeType]: (s: WhatShape<K>) => void
}
/* type ShapeMap = {
    0: (s: Circle) => void;
    1: (s: Square) => void;
} // same as before */

Playground link to code