John Maddock wrote:
Howard Hinnant wrote:
On Jul 25, 2006, at 1:45 PM, Howard Hinnant wrote:
namespace Mine {
struct Person {};
struct Female : public Person {};
bool isnormal(const Person&);
}
int main() { using namespace boost; // for boost::bind (just as an example) Mine::Female Jane; bool b = isnormal(Jane); } If this code doesn't make you nervous (because of the using directive), the related code below might:
namespace Mine {
struct sense_of_humor {};
struct Person : private boost::optional<sense_of_humor> {};
struct Female : public Person {};
bool isnormal(const Person&);
bool foo() { Female Jane; return isnormal(Jane); }
} // Mine
int main() { Mine::foo(); }
Again, isnormal will get hijacked by boost::isnormal(T) if it is in the translation unit (and in namespace boost and not somehow constrained).
Won't bool isnormal(const Person&) be the better overload and therefore the one called? The worst I can see happening is an "ambiguous overload" error.
BTW "isnorm" (note the spelling) should be in boost::math:: not boost:: so the risk is somewhat reduced.
template <class T> typename enable_if < std::numeric_limits<T>::is_specialized, bool
type isnormal(T t) {...}
Sigh, yes I can see the point of that, I'm fighting against it because it would break *my* code: it uses numeric types (NTL::RR) for which std::numeric_limits<> support is not appropriate, 'cos the precision is not a compile time constant. I could use another traits class that defaults to std::numeric_limits<T>::is_specialized but provides a backdoor for other types I guess.
John.
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The functions isnorm etc do not have to be in namespace boost. Put them in some nested namespace, such as boost::fp_classification. Then nobody is likely to bring them into scope unknowingly. --Johan Råde