フリー問題

問題文

Where can code call a private helper that an interface declares?

選択肢

1. Such helpers are visible only inside the declaring source, and they consequently cannot be reached from sibling routines under any condition.
2. Such helpers are visible only inside the declaring source, and they consequently cannot be reached from outer types under any condition.
3. Such helpers are visible only inside the open range, and they consequently cannot be reached from sibling routines under any condition.
4. Such helpers are visible only inside the open range, and they consequently cannot be reached from outer types under any condition.