为什么 typescript 类型保护不适用于 object 的内部属性

Question

Using typescript, when a function returns an object that has properties that might be null. Why does using a typeguard on these inner properties not allow typescript to infer that the inner prop must not be null after the guard?

Here's a minimum example of this. Try it

interface DatabaseResponse {
  settings: string | null
}
interface MainResponse {
  settings: string
}

const retrieveFromDatabase = (): DatabaseResponse => {
  return {
    settings: 'always a string but lets pretend it could be null sometimes'
  }
}

const main = (): MainResponse | Error => {
  const data = retrieveFromDatabase()

  if (data.settings === null) {
    throw new Error()
  }

  return data

}

The error for the return of the main function is

Type 'DatabaseResponse' is not assignable to type 'MainResponse | Error'.
  Type 'DatabaseResponse' is not assignable to type 'MainResponse'.
    Types of property 'settings' are incompatible.
      Type 'string | null' is not assignable to type 'string'.
        Type 'null' is not assignable to type 'string'.

Answer 1

That's not really a type guard. You can do this instead:

interface DatabaseResponse {
    settings: string | null
}
interface MainResponse {
    settings: string
}

const retrieveFromDatabase = (): DatabaseResponse => {
    return {
        settings: 'always a string but lets pretend it could be null sometimes'
    } as DatabaseResponse
}

const main = (): MainResponse | Error => {
    const data = retrieveFromDatabase()

    if (!isMainResponse(data)) {
        throw new Error()
    }

    return data
}

const isMainResponse = (data: DatabaseResponse | MainResponse): data is MainResponse {
  return !!data.settings;
}

Or similarly, create a new nonNullable type with the same typeguard

type NonNullableDB = { [K in keyof DatabaseResponse]: NonNullable<DatabaseResponse[K]> }

const isMainResponse = (data: DatabaseResponse | MainResponse): data is NonNullableDB => {
    return !!data.settings;
}