I'm reading a book. Obviously there are some failures in it.
In the code below, how can I access plant::operator<< in run() and plant::debug::operator<< in diagnostic()? The compiler can't decide between multiple operator<< in method diagnostic(). How can I tell it which one to use?
#include <iostream>
#include <string>
namespace plant {
class Tree
{
std::string name_;
public:
explicit Tree(const std::string_view name) : name_{name} {}
void print(std::ostream& os) const { os << name_; }
};
std::ostream& operator<<(std::ostream& os, const Tree& arg)
{
arg.print(os); return os;
}
namespace examples {
}
namespace debug {
std::ostream& operator<<(std::ostream& os, const Tree& arg)
{
os << "DEBUG: .."; arg.print(os); return os;
}
}
}
plant::Tree baum{"Mein Baum"};
void run()
{
using namespace plant;
std::cout << baum << "\n";
}
void diagnostic()
{
using namespace plant::debug;
std::cout << baum << "\n";
}
int main() {
run();
diagnostic();
}
Inside of diagnostic(), both operator<<'s are found as candidates, hence the ambiguity. plant::operator<< is found due to Argument-Dependent Lookup (which makes the using in run() redundant), and plant::debug::operator<< is found due to its using statement.
To resolve the ambiguity, you could call the debug::operator<< directly, eg:
void diagnostic()
{
//using namespace plant::debug;
plant::debug::operator<<(std::cout, baum) << "\n";
}
However, that kind of defeats the purpose of defining an operator<< in the first place, rather than a standalone function.
One option is you can instead simply move plant::operator<< into its own namespace, and then update run() to use that namespace, eg:
namespace plant {
class Tree
{
...
};
namespace no_debug {
std::ostream& operator<<(std::ostream& os, const Tree& arg)
{
...
return os;
}
}
namespace debug {
std::ostream& operator<<(std::ostream& os, const Tree& arg)
{
os << "DEBUG: ..";
...
return os;
}
}
}
...
void run()
{
using namespace plant::no_debug;
std::cout << baum << "\n";
}
void diagnostic()
{
using namespace plant::debug;
std::cout << baum << "\n";
}