Function declarations inside function expressions should inherit control flow narrowings of the parent function expression

I am trying to avoid typing null checks for properties that are formerly optional on the outer interface, but are then assigned with default values for the whole context of the function body.

I used a type guard for it, the problem is that the guarded type is lost in a function, but not lost in an arrow function, even though it can't be hoisted before the type guard.

I don't want to introduce new variables for every function parameter in a similar manner and I don't want to call other functions with casting as printMenu(options as DefinedOptions).

TypeScript Version: 3.7.3, 3.8.0-beta

Search Terms:
typescript type guard function arrow function
Code

export interface Options {
  header?: string
  border?: boolean
  pageSize?: number
  helpMessage?: string
  showKeypress?: boolean
}

interface DefinedOptions extends Options {
  pageSize: number
  helpMessage: string
}

const isType = <T>(arg: any): arg is T => true


export default async function menu(options: Options) {

  options // Options

  options.pageSize = options.pageSize ?? 0
  options.helpMessage = options.helpMessage ?? 'Default message'
  if (!isType<DefinedOptions>(options)) return null


  options // DefinedOptions


  return new Promise((resolve, reject) => {

    function handleMenuKeypress(key: any) {
      options // Options  -  should be DefinedOptions

      printMenu(options) // error
    }

    const candleMenuKeypress = (key: any) => {
      options // DefinedOptions

      printMenu(options) // no error
    }
  })
}

function printMenu(options: DefinedOptions) {

}

Playground Link

Expected behavior:

Type should be DefinedOptions in both.

Actual behavior:

Type is Options in function but DefinedOptions in arrow function.

Related:
#10927

Riproduci l’esempio nel TypeScript Playground collegato usando le versioni di TypeScript riportate e confronta i tipi inferiti per la funzione annidata e la funzione freccia. Il lavoro è completato quando entrambe le forme mantengono il narrowing di DefinedOptions dell’espressione di funzione padre senza cast o variabili aggiuntive.