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?
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.
Foo::Qux only contains the literal it already serves the purpose of separating the literals from the rest. It could be renamed into literals though.
Qux has other members that just aren't shown here.
Foo::Qux::operator_quux (100)but that didn't seem to work.operator<<forBaroperator ""in the wild before. Saw it had been added in C++11, but never looked into it. Now that I have, that's a <expletive deleted> to find documentation on. cppreference doesn't even seem to cover it.Foo::Qux::operator""_quux, notFoo::Qux::operator_quux. You can invoke it with that, or you can also dousing Foo::Qux::operator""_quux;to use the literal operator withoutusingthe rest of the namespace.