如何做这种类型-安全：“类型'Vehicle<C>‘和'Vehicle<CarConfig>’没有重叠”

Question

这个代码(这很奇怪，但我这么做只是为了提出这个问题)：

interface Vehicle<C> {
    id: string
    name: string
    config: C | null
}

interface CarConfig {
    leatherSeats: boolean
}

const Car: Vehicle<CarConfig> = {
    id: "car",
    name: "car",
    config: null
}

interface TruckConfig {
    numberOfWheels: number
}

const Truck: Vehicle<TruckConfig> = {
    id: "truck",
    name: "truck",
    config: null
}

function getVehicleConfig<C>(vehicle: Vehicle<C>): C | null {
    if (vehicle === Car) {
        return vehicle.config
    }
    if (vehicle === Truck) {
        return vehicle.config
    }
    throw new Error("Unknown vehicle")
}

使用此错误进行编译：

src/explain.ts:28:9 - error TS2367: This condition will always return 'false' since the types 'Vehicle<C>' and 'Vehicle<CarConfig>' have no overlap.

这里的期望是(由于软件的另一部分中的约束)，getVehicleConfig中的getVehicleConfig总是Car对象或Truck对象，getVehicleConfig将返回与实例关联的配置。

以一种类型安全的方式表达getVehicleConfig的正确方法是什么？

Answer 1

要走的道路是一个受歧视的工会：

type VehiculeType= 'Car' | 'Truck';


interface Vehicle<C> {
    id: string
    name: string
    config: C | null
    type: VehiculeType;
}

interface CarConfig {
    leatherSeats: boolean
}

const Car: Vehicle<CarConfig> = {
    type: 'Car',
    id: "car",
    name: "car",
    config: null
}

interface TruckConfig {
    numberOfWheels: number
}

const Truck: Vehicle<TruckConfig> = {
    type: 'Truck',
    id: "truck",
    name: "truck",
    config: null
}

function getVehicleConfig<C>(vehicle: Vehicle<C>): C | null {
    if (vehicle.type === 'Car') {
        return vehicle.config
    }
    if (vehicle.type === 'Truck') {
        return vehicle.config
    }
    throw new Error("Unknown vehicle") 
}