boost::serialization - Non intrusive serialization of user-defined members?
Hi, I have a class structure like class MemberClass; class ContainerClass { MemberClass special_member; int other_members; }; Following the example in http://www.boost.org/doc/libs/1_66_0/libs/serialization/doc/ i want to write a non-intrusive serialization: template<class Archive> void serialize(Archive & ar, ContainerClass & c, const unsigned int version) { ar & c.other_members; // easy - built-in type serialize(ar, c.special_member, version); //use same version??? } Assuming there's is also a non-intrusive void serialize(Archive & ar, MemberClass & m, const unsigned int version); this compiles and works but i feel this is not the way to do it because it reuses the version number of the container class for the member class? So how to handle that? Must the Container class manage version of its members individually? I suggest to add a section "Non-intrusively Serializable Members" to documentation with an example. Cheers, Lars R.
On 3/5/18 5:12 AM, Lars Ruoff via Boost-users wrote:
Hi, I have a class structure like
class MemberClass;
class ContainerClass { MemberClass special_member; int other_members; };
Following the example in http://www.boost.org/doc/libs/1_66_0/libs/serialization/doc/ i want to write a non-intrusive serialization:
template<class Archive> void serialize(Archive & ar, ContainerClass & c, const unsigned int version) { ar & c.other_members; // easy - built-in type serialize(ar, c.special_member, version); //use same version??? }
Assuming there's is also a non-intrusive void serialize(Archive & ar, MemberClass & m, const unsigned int version);
this compiles and works but i feel this is not the way to do it because it reuses the version number of the container class for the member class? So how to handle that? Must the Container class manage version of its members individually? I suggest to add a section "Non-intrusively Serializable Members" to documentation with an example.
Cheers, Lars R.
I'm not sure what the "special member" of the container class is. Presumably it's your own class as STL classes have no "special members". Assuming this to be the case there should be something like: struct ContainerClass { ... void serialize(Achive &ar, const unsigned int v){ ar & other_members; ar & special_member } ... }; or template<class Archive> void serialize(Archive & ar, ContainerClass & c, const unsigned int version) { ar & c.other_members; // easy - built-in type ar & c.special_member; }
Maybe i didn't express it clearly enough. I want non-intrusive serialization. I want to modify neither ContqinerClass, nor MemberClass. (Both are just custom classes, no link with STL containers. I just wanted to express that Container contains an instance of Member) So how can i write ar & c.special_member; since there is no serialize member function for this class? And when i'll be using an non-intrusive (free function) serialize, how do i handle the version number? On Mon, Mar 5, 2018 at 4:51 PM, Robert Ramey via Boost-users < boost-users@lists.boost.org> wrote:
On 3/5/18 5:12 AM, Lars Ruoff via Boost-users wrote:
Hi, I have a class structure like
class MemberClass;
class ContainerClass { MemberClass special_member; int other_members; };
Following the example in http://www.boost.org/doc/libs/ 1_66_0/libs/serialization/doc/ i want to write a non-intrusive serialization:
template<class Archive> void serialize(Archive & ar, ContainerClass & c, const unsigned int version) { ar & c.other_members; // easy - built-in type serialize(ar, c.special_member, version); //use same version??? }
Assuming there's is also a non-intrusive void serialize(Archive & ar, MemberClass & m, const unsigned int version);
this compiles and works but i feel this is not the way to do it because it reuses the version number of the container class for the member class? So how to handle that? Must the Container class manage version of its members individually? I suggest to add a section "Non-intrusively Serializable Members" to documentation with an example.
Cheers, Lars R.
I'm not sure what the "special member" of the container class is. Presumably it's your own class as STL classes have no "special members". Assuming this to be the case there should be something like:
struct ContainerClass { ... void serialize(Achive &ar, const unsigned int v){ ar & other_members; ar & special_member } ... };
or
template<class Archive> void serialize(Archive & ar, ContainerClass & c, const unsigned int version) { ar & c.other_members; // easy - built-in type ar & c.special_member; }
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non-intrusive serialisation is covered here: http://www.boost.org/doc/libs/1_66_0/libs/serialization/doc/serializati on.html#splittingfreefunctions given classes: MyNamespace::MyClass MyNamespace::MyContainer you could write: namespace boost { namespace serialization { template<class Archive> void serialize(Archive & ar, MyNamespace::MyContainer& c, const unsigned int version) { ar & c.other_members; // easy - built-in type ar & c.special_member; } template<class Archive> void serialize(Archive & ar, MyNamespace::MyClass& c, const unsigned int version) { ar & c.foo; // easy - built-in type if (version > 1) { ar & c.bar; // easy - built-in type } } }} Or you could put the serialize template functions in the MyNamespace namespace and reply on ADL. On Mon, 2018-03-05 at 17:36 +0100, Lars Ruoff via Boost-users wrote:
Maybe i didn't express it clearly enough. I want non-intrusive serialization. I want to modify neither ContqinerClass, nor MemberClass. (Both are just custom classes, no link with STL containers. I just wanted to express that Container contains an instance of Member)
So how can i write ar & c.special_member; since there is no serialize member function for this class?
And when i'll be using an non-intrusive (free function) serialize, how do i handle the version number?
On Mon, Mar 5, 2018 at 4:51 PM, Robert Ramey via Boost-users <boost-u sers@lists.boost.org> wrote:
On 3/5/18 5:12 AM, Lars Ruoff via Boost-users wrote:
Hi, I have a class structure like
class MemberClass;
class ContainerClass { MemberClass special_member; int other_members; };
Following the example in http://www.boost.org/doc/libs/1_66_0/lib s/serialization/doc/ i want to write a non-intrusive serialization:
template<class Archive> void serialize(Archive & ar, ContainerClass & c, const unsigned int version) { ar & c.other_members; // easy - built-in type serialize(ar, c.special_member, version); //use same version??? }
Assuming there's is also a non-intrusive void serialize(Archive & ar, MemberClass & m, const unsigned int version);
this compiles and works but i feel this is not the way to do it because it reuses the version number of the container class for the member class? So how to handle that? Must the Container class manage version of its members individually? I suggest to add a section "Non-intrusively Serializable Members" to documentation with an example.
Cheers, Lars R.
I'm not sure what the "special member" of the container class is. Presumably it's your own class as STL classes have no "special members". Assuming this to be the case there should be something like:
struct ContainerClass { ... void serialize(Achive &ar, const unsigned int v){ ar & other_members; ar & special_member } ... };
or
template<class Archive> void serialize(Archive & ar, ContainerClass & c, const unsigned int version) { ar & c.other_members; // easy - built-in type ar & c.special_member; }
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On 3/5/18 8:36 AM, Lars Ruoff via Boost-users wrote:
Maybe i didn't express it clearly enough. I want non-intrusive serialization. I want to modify neither ContqinerClass, nor MemberClass. (Both are just custom classes, no link with STL containers. I just wanted to express that Container contains an instance of Member)
So how can i write ar & c.special_member; since there is no serialize member function for this class?
That's the problem. the type of c.special_member must be serializable. If it's not already, you have to make it serializable by writing your own serialize function for it. It can be either intrusive or non-intrusive - doesn't matter. But it has to be there. The versioning of type type of special_member is handled in the serialization of that type.
Sorry, still don't understand. As said, i have defined void serialize(Archive & ar, MemberClass & m, const unsigned int version); so template<class Archive> void serialize(Archive & ar, ContainerClass & c, const unsigned int version) { ar & c.other_members; // easy - built-in type serialize(ar, c.special_member, version); //use same version??? } compiles. My question is what *version* number do i need to provide in the call to serialize for c.special_member? - The same as with what serialize(Archive & ar, ContainerClass & c, const unsigned int version) got called? (Like i did in example) This doesn't make sense to me. - Or I need to handle it myself, say i need to call serialize(ar, c.special_member, 3); because i *know* the current version of MemberClass is version 3. - Or there is some archive magic i don't see. On Mon, Mar 5, 2018 at 5:52 PM, Robert Ramey via Boost-users < boost-users@lists.boost.org> wrote:
On 3/5/18 8:36 AM, Lars Ruoff via Boost-users wrote:
Maybe i didn't express it clearly enough. I want non-intrusive serialization. I want to modify neither ContqinerClass, nor MemberClass. (Both are just custom classes, no link with STL containers. I just wanted to express that Container contains an instance of Member)
So how can i write ar & c.special_member; since there is no serialize member function for this class?
That's the problem. the type of c.special_member must be serializable. If it's not already, you have to make it serializable by writing your own serialize function for it. It can be either intrusive or non-intrusive - doesn't matter. But it has to be there. The versioning of type type of special_member is handled in the serialization of that type.
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On 3/5/18 9:06 AM, Lars Ruoff via Boost-users wrote:
Sorry, still don't understand. As said, i have defined void serialize(Archive & ar, MemberClass & m, const unsigned int version);
so template<class Archive> void serialize(Archive & ar, ContainerClass & c, const unsigned int version) { ar & c.other_members; // easy - built-in type serialize(ar, c.special_member, version); //use same version??? } compiles.
where is template<class Archive> void serialize(Archive & ar, special_member_type & t, const unsigned int version){ ar & ??; }; ? If you have this you could write: template<class Archive> void serialize(Archive & ar, ContainerClass & c, const unsigned int version){ ar & c.other_members; // easy - built-in type ar & c.special_member; //uses it's own version } Robert Ramey
Ah, ok. I assumed the syntax ar & c.special_member; //uses it's own version only worked with the intrusive version (i.e. serialize as a member of MemberClass). Seems it still works with non-intrusive. Pure magic. Thanks! On Mon, Mar 5, 2018 at 6:31 PM, Robert Ramey via Boost-users < boost-users@lists.boost.org> wrote:
On 3/5/18 9:06 AM, Lars Ruoff via Boost-users wrote:
Sorry, still don't understand. As said, i have defined void serialize(Archive & ar, MemberClass & m, const unsigned int version);
so template<class Archive> void serialize(Archive & ar, ContainerClass & c, const unsigned int version) { ar & c.other_members; // easy - built-in type serialize(ar, c.special_member, version); //use same version??? } compiles.
where is
template<class Archive> void serialize(Archive & ar, special_member_type & t, const unsigned int version){ ar & ??; };
?
If you have this you could write:
template<class Archive> void serialize(Archive & ar, ContainerClass & c, const unsigned int version){ ar & c.other_members; // easy - built-in type ar & c.special_member; //uses it's own version }
Robert Ramey
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participants (4)
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Richard Hodges -
Robert Ramey