4.4 regression: `in` operator does not narrow types in generic constraints when destructuring

[adalinesimonian]

Bug Report

🔎 Search Terms

in operator, narrowing, generics, constraints, destructuring

🕗 Version & Regression Information

• This changed between versions 4.3.5 and 4.4.2

⏯ Playground Link

Playground link with relevant code

💻 Code

type X = { a: string } | { b: string }

// Works in 4.3.5 and 4.4.2
function func1<T extends X>(value: T) {
  if ('a' in value) {
    const a = value.a
    console.log('a', a)
  } else {
    const b = value.b
    console.log('b', b)
  }
}

// Works in 4.3.5, not in 4.4.2
function func2<T extends X>(value: T) {
  if ('a' in value) {
    const { a } = value // Property 'a' does not exist on type 'X'.(2339)
    console.log('a', a)
  } else {
    const { b } = value // Property 'b' does not exist on type 'X'.(2339)
    console.log('b', b)
  }
}

🙁 Actual behavior

Types are not narrowed when using destructuring assignment and the compiler throws error 2339.

🙂 Expected behavior

Types are narrowed when using destructuring assignment.

[andrewbranch]

I’d start by making sure getNarrowableTypeForReference is returning X rather than T for the reference value, since that’s what 4.3 added that makes this possible in the first place.

[a-tarasyuk]

I’d start by making sure getNarrowableTypeForReference is returning X rather than T

@andrewbranch It seems this commit 0cc5c7b changed getNarrowableTypeForReference behavior to returning T rather than X for binding patterns.

[andrewbranch]

I subsequently killed more contextual typing by binding patterns in #45719, not sure at a glance whether that will play a role here or not