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4.4 regression: in operator does not narrow types in generic constraints when destructuring #45762
Description
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.
Activity
andrewbranch commentedon Sep 7, 2021
I’d start by making sure
getNarrowableTypeForReferenceis returningXrather thanTfor the referencevalue, since that’s what 4.3 added that makes this possible in the first place.a-tarasyuk commentedon Sep 12, 2021
Andrew Branch (@andrewbranch) It seems this commit 0cc5c7b changed
getNarrowableTypeForReferencebehavior to returningTrather thanXfor binding patterns.andrewbranch commentedon Sep 13, 2021
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