Binary expression in an `if` condition can't narrow identifier reassigned within expression of the optional chain

🔎 Search Terms

binary condition refinement predicate optional chain undefined

🕗 Version & Regression Information

• This is the behavior in every version I tried

⏯ Playground Link

https://www.typescriptlang.org/play?ts=5.4.0-dev.20240109#code/C4TwDgpgBAKu0F4oG8BQUoEsAmAuKAdgK4C2ARhAE4DcqAvraqJFAJIEBmVEBAxhOw4B7KEjQZmEAAoBDSjJIRgVfHEi0MmEmAA2mCNgCClTKAD8+YuSq0GqVNgi8dc6ByJ9gmIQSgBzJUFuPgFOIQAxIUo1CAAKSVV4AEp8IMoefkERAB8oD0cOTAIDRndPb18OAgBGeOrEyAAaKGAAJgaIJJR0KB0lLDDRf0DOYMywyOj4OqSNLA4oWKLhMwA6LV19IxNQKABCBCR8iELi7C7xDAHhde09A2NTEDm6KAgdAGdoTAXYpcGkAFgGkMqFhJMYvFWkkkmsNvdtk99oc8gQCkUDBcepowrdNg8ds8enR6PYyrwvD4oFVWnUOs02h0sRg+sBriJASMuOkQlkIdNgNVZj0fotlkI4Xcto9dgcjmiThjzt0ruy8QiZUSMK93l95ot-sIhkCQbyJlFIW0YZL8YjQMyruL1dLCS9SUA

💻 Code

type Type = {
  id: number;
};

type InferenceInfo = {
  typeParameter: Type;
  impliedArity?: number;
};

declare function getInferenceInfoForType(type: Type): InferenceInfo | undefined;

function fn1(t1: Type, t2: Type) {
  let info = getInferenceInfoForType(t1);
  if (info?.impliedArity !== undefined) {
    info.impliedArity; // ok
  } else if ((info = getInferenceInfoForType(t2))?.impliedArity !== undefined) {
    info.impliedArity; // 'info' is possibly 'undefined'.(18048)
  }
}

function fn2(t1: Type, t2: Type) {
  let info = getInferenceInfoForType(t1);
  if (info?.impliedArity !== undefined) {
    info.impliedArity; // ok
  } else if ((info = getInferenceInfoForType(t2))?.impliedArity) {
    info.impliedArity; // ok
  }
}

🙁 Actual behavior

The second branch in fn1 contains an error

🙂 Expected behavior

I'd expect this to be narrowed just fine - like the first branch and the second branch in fn2

Additional information about the issue

No response

リンクされた TypeScript Playground の再現コードから始め、fn1 と fn2 の narrowing の挙動を比較します。特に、再代入された optional chaining 式に注目してください。2 番目の分岐の型チェックの経路を追跡します。fn1 が info.impliedArity へのアクセスをエラーなしで受け入れ、既存のチェックを維持できれば完了です。