boost::format and custom printing a std containers
boost, boost-format, c++, namespaces
Solution
I think the most clean way is to provide a thin wrapper in your own namespace for each of the operators you want to override. For your case, it can be:
namespace ns
{
namespace wrappers
{
template<class T>
struct out
{
const std::set<T> &set;
out(const std::set<T> &set) : set(set) {}
friend std::ostream& operator<<(std::ostream& stream, const out &o)
{
stream << "{";
bool first = true;
for (const T& item : o.set)
{
if (!first)
stream << ", ";
else
first = false;
stream << item;
}
stream << "}";
return stream;
}
};
}
template<class T>
wrappers::out<T> out(const std::set<T> &set)
{
return wrappers::out<T>(set);
}
}
Then use it like this:
std::cout << boost::format("%1%") % ns::out(x);
Problem
I have a function in my namespace `ns` that helps me print STL containers. For example: ``` template <typename T> std::ostream& operator<<(std::ostream& stream, const std::set<T>& set) { stream << "{"; bool first = true; for (const T& item : set) { if (!first) stream << ", "; else first = false; stream << item; } stream << "}"; return stream; } ``` This works great for printing with `operator <<` directly: ``` std::set<std::string> x = { "1", "2", "3", "4" }; std::cout << x << std::endl; ``` However, using `boost::format` is impossible: ``` std::set<std::string> x = { "1", "2", "3", "4" }; boost::format("%1%") % x; ``` The problem is fairly obvious: Boost has no idea that I would like it to use my custom `operator <<` to print types which have nothing to do with my namespace. Outside of adding a `using` declaration into `boost/format/feed_args.hpp`, is there a convenient way to make `boost::format` look for my `operator <<`?