} // int a and float b ends there they not just end at the same time, but in reverse order of starting the lifetime.   You can do   std::string s{"xyz"}; std::string_view v{s};   First the string_view is destructed, then the string. As if there would be an outer and an inner block. They are not destructed at the same time.     -----Ursprüngliche Nachricht----- Von:SD SH via Std-Proposals <std-proposals_at_[hidden]> Gesendet:Fr 12.12.2025 15:54 Betreff:[std-proposals] 回复: 回复: [PXXXXR0] Add a New Keyword ‘undecl’ An:std-proposals_at_[hidden]; CC:SD SH <Z5515zwy_at_[hidden]>; Thanks for your dicussion in your example, {     int a; // int a starts there     float b;  // float b starts there } // int a and float b ends there {     int a; // int a starts there     { // scope A         float b; // float b starts there         using b = const; // OK, it covers the lifeline of 'b'         using a = const; // OK, it just make 'a' be const // it is allowed but I didn't mentioned above         using a = float { a }; // error, because float a is starts in scope A and the const int a couldn't end there     } // float b ends there; if "using a = float { a };" is allowed, both of the lifeline of const int a and float a will end there, but actually it is unexcepted     using a = float { a }; // int a ends there and float a starts there } // float a ends there  --------------------------------