フリー問題

問題文

A sealed interface permits two record classes, and a switch expression over that interface omits the default label. Why does the code compile?

選択肢

1. The compiler establishes exhaustiveness from the permitted subtypes, so it considers the absent fallback branch defective in this situation.
2. The compiler establishes exhaustiveness from the permitted subtypes, so it considers the absent fallback branch permissible in this situation.
3. The compiler establishes exhaustiveness from the loaded subtypes, so it considers the absent fallback branch defective in this situation.
4. The compiler establishes exhaustiveness from the loaded subtypes, so it considers the absent fallback branch permissible in this situation.