Hi! Why do you need an enable_if if you exactly know the type T? In that case you can use the temmplate specialization. template<> void f(const std::string&, bool) {...} template<> void f(const std::string&, foo) {...} template<typename T> void f(const std::string&, typename enable_if<boost::is_enum<T>, T>::type t) {...} On Sat, May 23, 2009 at 7:19 AM, Zachary Turner <divisortheory@gmail.com>wrote:
Suppose I have a template function
template<typename T> void f(const std::string&, T);
which I want there to be a couple of different variants of.
Version 1 - Enabled if and only if boost::is_same<T, bool> Version 2 - Enabled if and only if boost::is_same<T, foo> (foo is a class) Version 3 - Enabled if and only if boost::is_enum<T> Version 4 - Otherwise
Is there an easy way to express this? It seems I run into a bunch of overload ambiguity problems. I'm sure I can get around this by adding to every version the negation of all the other conditions, but this seems unnecessarily complicated.
_______________________________________________ Boost-users mailing list Boost-users@lists.boost.org http://lists.boost.org/mailman/listinfo.cgi/boost-users