699. TypeScript Union Narrowing with in Operator

Medium•

Predict whether this compiles and why.

type Cat = { kind: 'cat'; meow: () => void }
type Dog = { kind: 'dog'; bark: () => void }

function speak(pet: Cat | Dog) {
  if ('meow' in pet) {
    pet.meow()
  } else {
    pet.bark()
  }
}

Which statement is correct?

Does not compile. The in operator cannot narrow unions.Does not compile because methods are function types.Compiles only if you check pet.kind instead of in.Compiles. The in operator narrows to Cat in the true branch and Dog in the else branch.

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