Does the C++11 standard guarantee “n2 is int&” by “auto n2 = const_cast<int&>(n);”?

Question

const int n = 0;
auto& n1 = const_cast<int&>(n);
auto n2 = const_cast<int&>(n);

Does the C++11 standard guarantee n2 is int& by auto n2 = const_cast<int&>(n); ?

Must I use auto& n1 = const_cast<int&>(n); instead of auto n2 = const_cast<int&>(n); ?

Are the two ways completely equivalent to each other as per the C++11 standard?

Answer 1

auto on its own never produces a reference type.

So n2 is an int type.

(If I had a dollar for every time I see code like for (auto s : expensive_deep_copy_container) ).

Answer 2

auto uses the same rules as regular function template argument deduction, which never deduces a reference.

C++14 decltype(auto) , on the other hand, can deduce a reference here. As well as C++11 auto&& .

const int n = 0;
auto a = const_cast<int&>(n);           // a is int
decltype(auto) b = const_cast<int&>(n); // b is int&
auto&& c = const_cast<int&>(n);         // c is int&