CODEWITHSUNDEEP
c++namespacesoperatorsliterals
cshastry
cshastry

Reputation: 43

How to use operators defined in namespaces?

I have an operator defined in a namespace as follows:

namespace Foo {
  class Bar {
    public:
      Bar(double val): baz(val) {}
    // Rest of my object here

    private:
      double baz;

  };

  namespace Qux {
    Bar operator ""
    _quux(long double opd) {
      return Bar(opd / 10);
    }
  }
}



int main() {
  using namespace Foo::Qux;
  std::cout << "100_quux" << std::endl;

}

How do I use the operator without introducing the Foo::Qux namespace into my main() scope?

Upvotes: 3

Views: 874

Answers (1)

NathanOliver
NathanOliver

Reputation: 180415

You can't qualify the namespace for user define literals like

std::cout << 100.0Foo::Qux::_quux << std::endl

But what you can do is use a using statement to import just the literal operator into main using

using Foo::Qux::operator""_quux;

and you would use it like

std::cout << 100.0_quux << std::endl;

You could also call the operator manually like

std::cout << Foo::Qux::operator""_quux(100.) << std::endl;

Another option would be to place your user define literals into a namespace called literals and then you can just import that into main. That would look like

namespace Foo {
  class Bar {
    public:
      Bar(double val): baz(val) {}
    // Rest of my object here

    private:
      double baz;

  };

  namespace Qux {
    inline namespace literals {
      Bar operator ""_quux(long double opd) {
        return Bar(opd / 10);
      }
    }
    // other Qux Members
  }
}



int main() {
  using namespace Foo::Qux::literals;
  std::cout << 100.0_quux << std::endl;

}

Note that literals is an inline namespace so that Qux can still access the members without additional qualification.

Upvotes: 13

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