在打字稿中，我不明白为什么我收到错误 'Service1 | Service2 | null' is not assignable to type '(Service1 & Service2) |无效的'

Question

我正在尝试为一个简单的个人应用程序创建一个“婴儿”IoC 容器。我试图用这段代码实现的是根据作为第一个参数

传递的键在方法registerService中为参数service定义类型

但我遇到了错误

Type 'Service1 | Service2 | null' is not assignable to type '(Service1 & Service2) | null'.
  Type 'Service1' is not assignable to type 'Service1 & Service2'.
    Property 'thing2' is missing in type 'Service1' but required in type 'Service2'.

我的代码：

export const SERVICE_ONE_KEY = 'service1'
export const SERVICE_TWO_KEY = 'service2'

type Service1 = {
  thing1: () => void
}

type Service2 = {
  thing2: () => void
}

type IocKeys = typeof SERVICE_ONE_KEY | typeof SERVICE_TWO_KEY
type Services = {
  [SERVICE_ONE_KEY]: Service1 | null
  [SERVICE_TWO_KEY]: Service2 | null
}

const registry: Services = {
  service1: null,
  service2: null
}

export const registerService = (name: IocKeys, service: Services[typeof name]): Services[typeof name] => {
  registry[name] = service // Error here
  return registry[name]
}

export const getService = (name: IocKeys): Services[typeof name] => {
    return registry[name]
}

可以尝试一个“有效”的例子here来了解问题所在！

更新 1

我还想根据 getService 方法

中的名称参数键定义一个 return 类型

Answer 1

这可以通过 generics:

来实现

export const registerService = <T extends IocKeys>(name: T, service: Services[T]): Services[T] => {
  registry[name] = service
  return registry[name]
}

export const getService = <T extends IocKeys>(name: T): Services[T] => {
  return registry[name]
}

Playground

在打字稿中，我不明白为什么我收到错误 'Service1 | Service2 | null' is not assignable to type '(Service1 & Service2) |无效的'

In typescript I don't understand why i getting the error 'Service1 | Service2 | null' is not assignable to type '(Service1 & Service2) | null'

typescript

typescript-typings