Re: [Boost-users] [Serialization] is_serializable trait?

I also took a stab at making a boostified version: #include <iostream> #include #include #include #include #include namespace boost { namespace detail { template <class T> struct has_serialize_mem_fun { typedef void (T::*SerializationFun)(int &, unsigned); template <SerializationFun> struct A {}; template <class Q> static ::boost::type_traits::yes_type has_serialize_mem_fun_tester(A<&Q::serialize>*); template <class Q> static ::boost::type_traits::no_type has_serialize_mem_fun_tester(...); BOOST_STATIC_CONSTANT(bool, value = (sizeof(has_serialize_mem_fun_tester<T>(0)) == sizeof(::boost::type_traits::yes_type))); }; template struct is_serializable_impl { BOOST_STATIC_CONSTANT(bool, value = has_serialize_mem_fun<T>::value); }; template <typename T> struct is_serializable_impl::type> { BOOST_STATIC_CONSTANT(bool, value = true); }; } BOOST_TT_AUX_BOOL_TRAIT_DEF1(is_serializable,T,::boost::detail::is_serializable_impl<T>::value) } struct A { template <class Stream> void serialize(Stream & strm, unsigned version) {} }; struct B {}; struct C { void serialize() {} }; int main() { using namespace std; cout << boolalpha << boost::is_serializable<A>::value << '\n' // true << boost::is_serializable<int>::value << '\n' // true << boost::is_serializable<float>::value << '\n' // true << boost::is_serializable<double>::value << '\n' // true << boost::is_serializable<char>::value << '\n' // true << boost::is_serializable<B>::value << '\n' // false << boost::is_serializable<C>::value << '\n' // false << boost::is_serializable<string>::value << endl; //false but should be true } Needs specializations for standard containers as well but this suits my needs currently. Thoughts? On 7/23/07, Chris Fairles wrote:

Thanks!

I kept trying things with void(A::*serialize)(blah, const unsigned int) in my test function. Like:

template<typename T> class IsSerializableClassT { private: typedef char One; typedef struct { char a[2]; } Two; template<typename C> static Two test(...); template<typename C> static One test(void(C::*)(int,int)); public: enum { Yes = sizeof(IsClassT<T>::template test<T>(0)) == 1 }; enum { No = !Yes }; };

But, when resolving test(...) vs. the member func pointer, it picks the mem func pointer for any class even if it doesn't have a func with that specific signature.

That local tempalte trick you use solves the issue. A local template with a non-type member-function parameter (which i only recently learned was even possible!), void(T::*)(int&,int)....

So the substitution fails if the supplied fuction doesn't match the signature. So whats the technical reason why the substitution doesn't fail based on mem func ptr signature in a function parameter? I guess function arguments aren't bound until runtime so it never really looks in the class for the function at compile time?

Chris

On 7/23/07, Robert Ramey wrote:

...

Very clever, I would expect some variation of this to end up somewhere in the library itself some day.

Robert Ramey

"Roman Perepelitsa" wrote in message news:OFC1A16EC6.69CEAA30-ONC3257321.002322F2-C3257321.0023748C@db.com...

...

Chris Fairles" on Fri, 20 Jul 2007 18:03:18 -0400 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?

You can try something like this.

#include <iostream> #include <iomanip>

typedef char yes_type;

struct no_type { char padding[8]; };

template <class T> struct HasSerializeMember { typedef void (T::*SerializationFun)(int &, unsigned);

template <SerializationFun> struct A {};

template <class Q> static yes_type Test(A<&Q::serialize>*);

template <class Q> static no_type Test(...);

static const bool value = sizeof(Test<T>(0)) == sizeof(yes_type); };

struct A { template <class Stream> void serialize(Stream & strm, unsigned version) {} };

struct B { };

struct C { void serialize() {} };

int main() { using namespace std; cout << boolalpha << HasSerializeMember<A>::value << '\n' // true << HasSerializeMember<B>::value << '\n' // false << HasSerializeMember<C>::value << endl; // false }

Regards, Roman Perepelitsa

Deutsche Bank Moscow +7 (495) 660-74-08

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