Typescript: discriminating worker

[xxRockOnxx]

This is more of a Typescript question but how do I strongly-type Worker?

Here's a minimal reproduction:

interface JobData {
  id: string
}

interface FooData extends JobData {
  foo: string
}

interface BarData extends JobData {
  bar: boolean
}

type FooWorker = Worker<FooData, string, 'foo'>
type BarWorker = Worker<BarData, boolean, 'bar'>

const fooBarWorker: FooWorker | BarWorker = new Worker('worker', async (job) => {
  switch (job.name) {
    case 'foo':
      console.log(job.data.foo) // Error: Property `foo` does not exists on type `JobData`
      return job.data.foo
    case 'bar':
      console.log(job.data.bar) // Error: Property `bar` does not exists on type `JobData`
      return job.data.bar
  }
})

Using & instead of | somehow works but only for one type of worker

const fooBarWorker: FooWorker & BarWorker = new Worker('worker', async (job) => {
  switch (job.name) {
    case 'foo':
      console.log(job.data.foo)
      return job.data.foo

    // Error: Type 'bar' is not comparable to type 'foo'
    case 'bar':
      console.log(job.data.bar)
      return job.data.bar
  }
})

Related: #1156
Minimal reproduction: https://codesandbox.io/s/typescript-playground-export-forked-cmr7k2?file=/index.ts