What is the deduced type of a constexpr?
c++, c++11, constexpr, overloading, templates
Solution
`N3337 [dcl.constexpr]/9:` A `constexpr` specifier used in an object declaration declares the object as `const`. [...]
Since you declared `k` as `constexpr`, it is also declared as `const`, so the `const int&` is selected in overload resolution.
Problem
``` #include <iostream> #include <string> void foo(int& k) { std::cout << "int&\n"; } void foo(int&& k) { std::cout << "int&&\n"; } void foo(const int& k) { std::cout << "const int&\n"; } void foo(const int&& k) { std::cout << "const int&&\n"; } int main() { static constexpr int k = 1; foo(k); foo(1); } ``` The output is: ``` const int& int&& ``` What exactly is a constexpr variable treated as? The overload for `foo` gives `const int&`. Edit: Moving on with constexpr being deduced as `const T&`; Why does a constexpr at class scope fail to be passed to a function taking universal reference?! ``` #include <type_traits> template <typename T> void goo(T&& k) { static_assert(std::is_same<decltype(k), const int&>::value, "k is const int&"); } class F { static constexpr int k = 1; public: void kk2 () { goo(k); } }; int main () { F a; a.kk2(); } ``` The above fails to compile giving error `undefined reference to F::k` However the below passes: ``` #include <type_traits> template <typename T> void goo(T&& k) { static_assert(std::is_same<decltype(k), const int&>::value, "k is const int&"); } int main() { static constexpr int k = 1; goo(k); } ```