`using` declaration for a user-defined literal operator

Question

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

E.g.,

#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.

@PiotrSkotnicki That, my friend, is an answer. (A correct answer, even!) — Kyle Strand, Aug 31 '15 at 20:23

score 6 · Accepted Answer · edited Jun 20 '20 at 09:12

6

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

edited Jun 20 '20 at 09:12

Community

1
1

answered Aug 31 '15 at 20:24

Piotr Skotnicki

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So `user-defined-string-literal` is the syntax element corresponding to `""_hr`? – Kyle Strand Aug 31 '15 at 20:47
@KyleStrand yes, `user-defined-string-literal` is the same as `string-literal identifier` which are `""` and `_hr` respectively – Piotr Skotnicki Sep 01 '15 at 06:18

`using` declaration for a user-defined literal operator

1 Answers1