'in' does not remove undefined from indexed value (with noUncheckedIndexedAccess)

🔎 Search Terms

in operator, noUncheckedIndexedAccess

🕗 Version & Regression Information

Tested 5.8.2 and 5.9.0-dev.20250310

⏯ Playground Link

https://www.typescriptlang.org/play/?noUncheckedIndexedAccess=true&ts=5.8.2#code/MYewdgzgLgBAhgJwQLhmArgWwEYFMEDaAujALwwECMANDAEy0DMRA3AFCiSwCWAJgB6oMOfGXrtuAMxgAKPvxjcw8JAEoYAbzYwdMTtBiC0WPAjGJC81tt36QAG1wA6eyADmM-qvYBfNkA

💻 Code

With an array:

const arr: number[] = [1, 2, 3];
const idx: number = 2;
if (idx in arr) {
    const x: number = arr[idx];
    console.log(x);
}

With a map/object:

const map: Record<string, number> = { a: 1 };
const key: string = "a";
if (key in map) {
    const x: number = map[key];
    console.log(x);
}

🙁 Actual behavior

Type 'number | undefined' is not assignable to type 'number'.
  Type 'undefined' is not assignable to type 'number'.

🙂 Expected behavior

Expected 'in' operator to guarantee that the result of indexing is not undefined in this branch.

Additional information about the issue

No response

noUncheckedIndexedAccess を使用し、array と Record の両方の例を含む、リンクされた TypeScript Playground の再現から始めてください。型チェッカーが in 分岐内でインデックスアクセスをどのように絞り込むかを追跡してください。完了の条件は、オプションの既存の動作を弱めることなく、両方の例でインデックス化された値を number に代入できることです。