Why doesn't TypeScript type guard 'in' narrows types to keyof types?

Question

Consider this code:

const obj = {
    a: 1,
    b: 2
}

let possibleKey: string = 'a'

if (possibleKey in obj) console.log(obj[possibleKey])

When possibleKey in obj is true, we know that possibleKey has type keyof typeof obj, right? Why doesn't TypeScript type system detects that and narrows down string to that type? Instead, it says:

Element implicitly has an 'any' type because expression of type 'string' can't be used to index type '{ a: number; b: number; }'.

Answer 1

Per the docs:

For a n in x expression, where n is a string literal or string literal type and x is a union type, the “true” branch narrows to types which have an optional or required property n, and the “false” branch narrows to types which have an optional or missing property n.

In other words, n in x narrows x, not n, and only for string literals or string literal types in union types. For that expression to work, you'd have to give the compiler more information, e.g. using a type assertion:

if (possibleKey in obj) {
  console.log(obj[<keyof typeof obj>possibleKey]);
}