On Sat, 29 Jan 2005 17:56:59 +0200, Yuval Ronen <ronen_yuval@yahoo.com> wrote:
Hi. I encountered the need to query a boost::variant to which type it currently contain. I'm talking about something like:
variant<int, string> v(3); assert(v.is_a<int>()); assert(!v.is_a<string>());
Take a look at get(), it's almost what you need. Unfortunately, it compiles even when the type cannot be inside the variant. This can be fixed with mpl::contains: template<typename T, typename Variant> bool variant_is_a(const Variant& v) { BOOST_MPL_ASSERT(( boost::mpl::contains<typename Variant::types,T> )); return boost::get<T>(&v) != 0; } There should be a version of get that works as variant_is_a above. IMHO, the current behaviour is more dangerous than that of variant_is_a, so get() should be changed and maybe a new function be created that keeps the current behaviour. Some code may break with this change with a compile error that would go away if the other version of get is used. Bruno
This can be done using the which() method, but its drawback is that I rely on the order of the types in the variant declaration rather on the type, which is really what I'm interested in.
I can implement the variant_is_a<>() feature using a visitor such as:
template <typename T> struct is_a_visitor : public boost::static_visitor<bool> { template <typename U> bool operator()(const U &) const { return false; }
bool operator()(const T &) const { return true; } };
template <typename T, class Variant> bool variant_is_a(const Variant &v) { return v.apply_visitor(is_a_visitor<T>()); }
but this way has a problem: calling variant_is_a<double>() on a variant of type variant<int, string> (no double for this variant) will simply return false, rather than just not compile, which is what I prefer. Making this not compile will require a new feature from variant: a boolean constant 'can_be'. With this, the variant_is_a() function could be written with additional line:
BOOST_STATIC_ASSERT(Variant::can_be<T>::value);
IMO, the is_a/can_be features are general and useful enough to be included in the library. Wouldn't you agree?
Yuval