Conditional function type

Question

Given the following scenario,

function doSomething({ id, name, onSelect, onUnselect }: TDoSomething) {
  if (onSelect) {
    // I want `onUnselect` to be defined here
    // ...
  }

  return { id, name }
} 

type TDoSomething =
  | IDoSomethingWithoutSelect
  | IDoSomethingWithSelect

interface IDoSomethingWithoutSelect {
  id: string
  name: string
}

interface IDoSomethingWithSelect {
  id: string
  name: string

  onSelect: (id: string) => void
  onUnselect: (id: string) => void
}

What I want to do is (conceptually?) simple: want to have onUnselect to necessarily be defined if onSelect is. In other words, if I pass onSelect, then this function must expect a onUnselect as well.

(A playground to you can be accessed here.)

Turns out my attempts were failures. I don't know exactly what I should do to hit success.

(This is something else I've tried to do)

Eugene Karataev · Accepted Answer · 2020-04-24 18:28:10Z

2

You can use in operator to narrow argument type to IDoSomethingWithSelect

function doSomething(arg: TDoSomething) {
  const { id, name } = arg;
  if ('onSelect' in arg) {
    arg.onUnselect(id);
  }

  return { id, name }
}

answered Apr 24, 2020 at 18:28

Eugene Karataev

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2 Comments

Guilherme Oderdenge Over a year ago

One side question, though. When I call doSomething({ onSelect: () => {} }) it doesn't require me to input onUnselect. Is there a way to overcome this?

Eugene Karataev Over a year ago

Your example should fail with compilation error Argument of type '{ onSelect: () => void; }' is not assignable to parameter of type 'TDoSomething'.

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