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

From: Chris Fairles (chris.fairles_at_[hidden])
Date: 2007-07-23 13:22:57

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 <ramey_at_[hidden]> wrote:
>
>
> Very clever, I would expect some variation of this to end up somewhere in
> the library itself some day.
>
> Robert Ramey
>
> "Roman Perepelitsa" <roman.perepelitsa_at_[hidden]> wrote in message
> news:OFC1A16EC6.69CEAA30-ONC3257321.002322F2-C3257321.0023748C_at_db.com...
>
> > Chris Fairles" <chris.fairles_at_[hidden]> 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
> serialize<archive::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 <class T,class F=void>
> > struct Q{enum{value=false};};
> >
> > template<class T> struct
> Q<T,void(T::serialize)(int,int)>{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
> [phone]
>
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