Spread operator mixed with destructuring assignment resulting in unsoundness

Bug Report

This might have been covered before but I am not sure how to best search for it, apologies if it is a duplicate - when destructuring assignment is used with a spread operator (see example below) TS seems to ignore the possibility that the variable could be assigned undefined.

🔎 Search Terms

spread, destructure, destructuring

🕗 Version & Regression Information

• This is the behavior in every version I tried (since 4.1.5 which is the oldest version on the playground with noUncheckedIndexedAccess available)

⏯ Playground Link

Playground link with relevant code

💻 Code

const [a] = [...(false ? (["a"] as const) : [])];
//     ^?
const b = false ? (["a"] as const) : [];
//    ^?
const c = [...b];
//    ^?
const [d] = c;
//     ^?

const [a_prime] = [...(false ? (["a"] as const) : [] as const)];
//     ^?

🙁 Actual behavior

a is typed as "a", while d and a_prime are typed as "a" | undefined". Note this only happens when noUncheckedIndexedAccess is enabled, otherwise all of them are typed as "a".

Note that d is the unrolled version of a which makes it more surprising that TS is typing them differently.

🙂 Expected behavior

They should all be typed "a" | undefined.

Inizia con l'esempio collegato di TypeScript Playground e confronta i tipi inferiti di a, d e a_prime con noUncheckedIndexedAccess abilitato. Traccia la gestione da parte del compilatore del destructuring degli array e delle espressioni spread; il lavoro è completato quando tutti e tre i casi inferiscono coerentemente "a" | undefined e viene aggiunta la copertura di regressione per l'esempio segnalato.