`using` declaration for a user-defined literal operator

Question

Is it possible to have a using declaration for the literals operator, operator "" ?

Eg,

#include <chrono>

namespace MyNamespace
{
  constexpr std::chrono::hours operator "" _hr(unsigned long long n){
    return std::chrono::hours{n};
  }

  // ... other stuff in the namespace ...
}

using MyNamespace::operator"";    // DOES NOT COMPILE!

int main()
{
  auto foo = 37_hr;
}

My work-around has been to put these operators in their own nested namespace called literals , which allows using namespace MyNamespace::literals; , but this seems somewhat inelegant, and I don't see why the using directive can't be used for operator functions in the same way as it can for any other functions or types within a namespace.

Answer 1

using MyNamespace::operator""_hr;
//                           ^^^

DEMO

Grammar reference:

 using-declaration: using typename (opt) nested-name-specifier unqualified-id ; using :: unqualified-id ; unqualified-id: identifier operator-function-id conversion-function-id literal-operator-id ~ class-name ~ decltype-specifier template-id literal-operator-id: operator string-literal identifier operator user-defined-string-literal