How can you determine (compile-time) that a class has a serialize function? struct foo { template <typename Archive> void serialize (Archive &, const unsigned int); }; "&foo::serialize" isn't a type and I can't figure out a way to make it a type (besides a typedef in the class) and also a constant expression. Tried to play around with this as well but theres no substitution failure here and static functions can't be constant expressions (commonnnn c++0x!!) struct serialize { template <typename T> static void value(T const&) { void(T::*f)(archive::xml_oarchive &, const unsigned int) = &T::template serializearchive::xml_oarchive; (void)f; } }; Basically, in the code below I want to see "GoodGood" printed out and the only thing I can change is Q. struct A{ void f(int,int){} }; struct B{}; template struct Q{enum{value=false};}; template<class T> struct Q{enum{value=true};}; int main(){ if(Q<A>::value) { cout << "Good";} if(!Q<B>::value){ cout << "Good";} } Help/hints/remotely possible? Thanks, Chris On 4/20/07, Robert Ramey wrote:

a type is serializable if its a primtive or it has a serialize function.

Robert Ramey

Sohail Somani wrote:

...

Hi,

I know there is the tracking level trait. Is that sufficient to be able to say that T is serializable?

struct T { T(some_stuff); private: some_non_default_constructing_members; };

In this case, I think tracking level will be set to track selectively but its not serializable as is.

Thanks,

Sohail

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