[serialization] Different behaviors between archive and its polymorphic variant

Maxime van Noppen

7 Aug 2010 7 Aug '10

1:54 p.m.

Hi, I'm trying to make a custom archive class that works either with the template flavor or via the polymorphic_archive. It compiles but produces a different behavior and I can't find why. Basically I did a trivial oarchive class that overloads (via save_override) how are treated NVP. When used with the template version it works, whereas with the polymorphic version it doesn't. The code is attached. There is a preprocessor switch to choose the template or polymorphic version (-DPOLY). $ g++ poly.cc -lboost_serialization && ./a.out 42 $ g++ -DPOLY poly.cc -lboost_serialization && ./a.out 42 This is with Boost 1.43 and g++ 4.5.0. I looked at various archive implementations but couldn't find any code that I could relate to this problem. Thanks, -- Maxime

Attachments:

poly.cc (text/x-c++src — 3.0 KB)

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Maxime van Noppen

18 Aug 18 Aug

7:50 a.m.

On 08/07/2010 03:54 PM, Maxime van Noppen wrote:

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I looked at various archive implementations but couldn't find any code that I could relate to this problem.

Little bump. I still get the same behavior with Boost 1.44 so it seems I'm doing something wrong but I really can't find what. :-/ Any idea? Thanks, -- Maxime

Robert Ramey

2:04 p.m.

New subject: [serialization] Different behaviors between archive and its polymorphic variant

make a small example which illustrates the problem and we'll take a look a it. It's not clear to me that one should necessarily expect that the output of an archive class should be identical to it's polymorphic variant. Robert Ramey Maxime van Noppen wrote:

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On 08/07/2010 03:54 PM, Maxime van Noppen wrote:

...
I looked at various archive implementations but couldn't find any code that I could relate to this problem.

Little bump. I still get the same behavior with Boost 1.44 so it seems I'm doing something wrong but I really can't find what. :-/ Any idea?

Thanks,

Maxime van Noppen

1:22 p.m.

On 08/18/2010 04:04 PM, Robert Ramey wrote:

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make a small example which illustrates the problem and we'll take a look a it.

In the first post I attached such an example.

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It's not clear to me that one should necessarily expect that the output of an archive class should be identical to it's polymorphic variant.

So code that uses the polymorphic variant might not work with code that doesn't even though they use the same archive class though trough different methods ? It seems a bit strange to me. However, the problem isn't there actually. What happens in the example is that when used through the template variant, save_override is correctly used, though with the polymorphic one it isn't. -- Maxime

Robert Ramey

4:42 p.m.

New subject: [serialization] Different behaviors between archive and its polymorphic variant

Maxime van Noppen wrote:

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On 08/18/2010 04:04 PM, Robert Ramey wrote:

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make a small example which illustrates the problem and we'll take a look a it.

In the first post I attached such an example.

I went back and looked at your example. You're on the wrong track here. First, classes to be serialized should be not dependent in anyway upon any archive class. So use: class unit { public: unit() : id_(0) { } unit(int id) : id_(id) { } template <typename Archive> void serialize(Archive& ar, const unsigned int version) { ar & BOOST_SERIALIZATION_NVP(id_); } private: int id_; }; note that "polymorphism" is not implemented via virtual functions in the archive class itself so your archive should be dependent on whether or note you're using the polymorphic_archive template. class test_oarchive : public boost::archive::detail::common_oarchive { friend class boost::archive::save_access; friend class boost::archive::detail::interface_oarchive; public: test_oarchive(std::ostream& out, unsigned int = 0) : out_(out){ } void save_binary(const void* address, std::size_t count) { out_.write(static_cast(address), count); } template <typename T> void save(const T& t){ out_ << t; } template <class T> void save_override(const boost::serialization::nvp<T>& nvp, int f) { out_ << ""; typedef boost::archive::detail::common_oarchive detail_common_oarchive; this->detail_common_oarchive::save_override(nvp.const_value(), f); out_ << "\n"; } void save_override(const boost::archive::object_id_type & t, int) { } void save_override(const boost::archive::object_reference_type & t, int) { } void save_override(const boost::archive::version_type & t, int) { } void save_override(const boost::archive::class_id_type & t, int) { } void save_override(const boost::archive::class_id_optional_type & t, int) { } void save_override(const boost::archive::class_id_reference_type & t, int) { } void save_override(const boost::archive::class_name_type & t, int) { } void save_override(const boost::archive::tracking_type & t, int) { } private: std::ostream& out_; }; Test the above to verify that it works according to your taste. Now use a "templated typedef" to generate the polymorphic version: #include #include #include typedef boost::archive::detail::polymorphic_oarchive_route< test_oarchive

...

polymorphic_test_oarchive;

// required by export BOOST_SERIALIZATION_REGISTER_ARCHIVE( test_oarchive ) or something like this. Now you should have test_oarchive and polymorphic_test_oarchive Robert Ramey

Maxime van Noppen

4:10 p.m.

On 08/18/2010 06:42 PM, Robert Ramey wrote:

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I went back and looked at your example. You're on the wrong track here.

Thank you very much!

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First, classes to be serialized should be not dependent in anyway upon any archive class. So use:

Ouch. The reason why I wanted polymorphic archives was to be able to have a non-template serialize() method so that it may be implemented and compiled once for all in the .cpp file.

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Test the above to verify that it works according to your taste.

I still have a different behavior when serializing through polymorphic_test_oarchive.

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Now use a "templated typedef" to generate the polymorphic version:

#include #include #include

typedef boost::archive::detail::polymorphic_oarchive_route< test_oarchive

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polymorphic_test_oarchive;

// required by export BOOST_SERIALIZATION_REGISTER_ARCHIVE( test_oarchive )

or something like this. Now you should have test_oarchive and polymorphic_test_oarchive

That's what I'm doing. So now 'unit' only has a template serialize method, but I still

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int main() { unit u(42);

{ std::cout << "template:" << std::endl; test_oarchive ar(std::cout); ar << BOOST_SERIALIZATION_NVP(u); std::cout << std::endl; }

{ std::cout << "polymorphic:" << std::endl; polymorphic_test_oarchive ar(std::cout); ar << BOOST_SERIALIZATION_NVP(u); std::cout << std::endl; } }

Which outputs:

...

template: 42

polymorphic: 42

Code is attached. -- Maxime

Robert Ramey

6:33 p.m.

New subject: [serialization] Different behaviors between archive and its polymorphic variant

...

...
So now 'unit' only has a template serialize method, but I still

...
int main() { unit u(42);

{ std::cout << "template:" << std::endl; test_oarchive ar(std::cout); ar << BOOST_SERIALIZATION_NVP(u); std::cout << std::endl; }

{ std::cout << "polymorphic:" << std::endl; polymorphic_test_oarchive ar(std::cout); ar << BOOST_SERIALIZATION_NVP(u); std::cout << std::endl; } }

Which outputs:

...
template: 42

polymorphic: 42

Code is attached.

what happens when you try: std::cout << "polymorphic:" << std::endl; polymorphic_oarchive & po = polymorphic_test_oarchive ar(std::cout); po << BOOST_SERIALIZATION_NVP(u); std::cout << std::endl; This would actually be using the polymorphic interface as it's intended to be used by routing the serialization through the polymorphic interface. Robert Ramey

Maxime van Noppen

19 Aug 19 Aug

8:52 a.m.

On 08/18/2010 08:33 PM, Robert Ramey wrote:

...

...
...
So now 'unit' only has a template serialize method, but I still

...
int main() { unit u(42);

{ std::cout << "template:" << std::endl; test_oarchive ar(std::cout); ar << BOOST_SERIALIZATION_NVP(u); std::cout << std::endl; }

{ std::cout << "polymorphic:" << std::endl; polymorphic_test_oarchive ar(std::cout); ar << BOOST_SERIALIZATION_NVP(u); std::cout << std::endl; } }

Which outputs:

...
template: 42

polymorphic: 42

Code is attached.

what happens when you try:

std::cout << "polymorphic:" << std::endl; polymorphic_oarchive & po = polymorphic_test_oarchive ar(std::cout); po << BOOST_SERIALIZATION_NVP(u); std::cout << std::endl;

This would actually be using the polymorphic interface as it's intended to be used by routing the serialization through the polymorphic interface.

Robert Ramey

_______________________________________________ Boost-users mailing list Boost-users@lists.boost.org http://lists.boost.org/mailman/listinfo.cgi/boost-users

I added:

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{ std::cout << "polymorphic with polymorphic_oarchive:" << std::endl; polymorphic_test_oarchive ar(std::cout); boost::archive::polymorphic_oarchive & po = ar; po << BOOST_SERIALIZATION_NVP(u); std::cout << std::endl; }

The output is:

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template: 42

polymorphic: 42 polymorphic with polymorphic_oarchive: 42

-- Maxime

Dante Stroe

9:58 a.m.

New subject: [serialization] serialization of base pointer to derived class issue

Hi, I have a problem with serializing a derived classes through a pointer to a base class. I tried registering with the syntax from the documentation: |ar.template register_type(); | , which works well in case the derived class has a default constructor (just like the example). I just don't know what the syntax is for the case where the derived class does not have a default constructor (if I use the same syntax to register the derived class then I get a compilation error about the fact that there is no. I can provide an example in case this is not clear enough. Is it an absolute requirement for all derived classes to have a default constructor? Thank you and best regards, Dante

Robert Ramey

4:38 p.m.

New subject: [serialization] serialization of base pointer toderived class issue

Look at the documentation regarding LoadConstruct and also the tests for no default ctor. Robert Ramey Dante Stroe wrote:

...

Hi,

I have a problem with serializing a derived classes through a pointer to a base class. I tried registering with the syntax from the documentation:

ar.template register_type();

, which works well in case the derived class has a default constructor (just like the example). I just don't know what the syntax is for the case where the derived class does not have a default constructor (if I use the same syntax to register the derived class then I get a compilation error about the fact that there is no. I can provide an example in case this is not clear enough.

Is it an absolute requirement for all derived classes to have a default constructor?

Thank you and best regards, Dante

_______________________________________________ Boost-users mailing list Boost-users@lists.boost.org http://lists.boost.org/mailman/listinfo.cgi/boost-users

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Dante Stroe
Maxime van Noppen
Robert Ramey