If you utilize structured bindings like so
auto [a, b, c] = std::make_tuple(1, 10.0, "string object"s);
then will the copies from the returned tuple be elided and the objects go straight into a, b and c or will the initializations be move constructions from the individual tuple elements? I doubt that this would cause a copy to happen but I am not sure about whether the description of mandatory copy elision in the standard handles this case.
As covered by this excellent answer, the declaration is equivalent to:
auto e = std::make_tuple(1, 10.0, "string object"s);
int& a = get<0>(e);
double& b = get<1>(e);
std::string& c = get<2>(e);
except that there isn't a name e. The get function, in this context, yields an lvalue reference to the selected item.
In C++17, auto name = prvalue; is defined to declare an object called name of type decltype(prvalue) initialized with the prvalue expression -- there's no intermediate temporary that was elidable as in previous versions.
In other words, the e declaration behaves exactly the same as:
std::tuple<int, double, std::string> e {1, 10.0, "string object"s};
and then a,b,c are references to the elements of that tuple.
Note: The above explanation corresponds to the latest C++17 draft sources; behaviour may change before C++17 is finalized of course.
