Edit Question - KnowledgeHub

Title Body

Update: See the full answer below. The short answer is no, not directly. You can create an indirect reference using std::reference_wrapper or accomplish the same effect more generally with pointers (but without the syntactic sugar and added safety of references). I ask because tuples make a convenient variadic storage unit in C++11. In theory it sounds reasonable for one element of a tuple to hold a reference to another element in the same tuple. (Replace "reference" with "pointer" and it works in practice.) The devil is the details of constructing such a tuple. Consider the following example: #include <tuple> #include <iostream> class A { public: A() : val(42) { } int val; }; class B { public: B(A &a) : _a(a) { } int val() { return _a.val; } private: A &_a; }; int main() { A a; B b(a); std::tuple<A, B> t1(a, b); a.val = 24; std::cout << std::get<0>(t1).val << "\n"; // 42 std::cout << std::get<1>(t1).val() << "\n"; // 24 return 0; } The second element in the tuple t1 references the automatic variable a instead of the first element in t1 . Is there any way to construct a tuple such that one element of the tuple could hold a reference to another element in the same tuple? I'm aware that you could sort of achieve this result by creating a tuple of references, like this: int main() { A a; B b(a); std::tuple<A &, B &> t2 = std::tie(a, b); a.val = 24; std::cout << std::get<0>(t2).val << "\n"; // 24 std::cout << std::get<1>(t2).val() << "\n"; // 24 return 0; } But for my purposes that's cheating, since the second element in t2 is still ultimately referencing an object that lives outside of the tuple. The only way I can think of doing it compiles fine but may cont

Tags (comma-separated) Cancel