提高序列化按位可序列化 [英] Boost serialization bitwise serializability

问题描述

我希望从 is_bitwise_serializable trait 序列化类如下(没有序列化函数):

I expect from is_bitwise_serializable trait to serialize class like following (without serialize function):

class A { int a; char b; };
BOOST_IS_BITWISE_SERIALIZABLE(A);
A a{2, 'x'};
some_archive << a; // serializes a bitwisely

我想知道，为什么需要为 bitwise_serializable 类提供序列化功能?

I wonder, why is there a need to provide serialize function for bitwise_serializable class?

推荐答案

来自文档:

一些简单的类可以通过直接复制类的所有位来序列化.对于不包含指针成员且既未进行版本控制也未进行跟踪的 POD 数据类型，尤其如此.一些档案，例如不可移植的二进制档案可以让我们利用这些信息大大加快序列化速度.

Some simple classes could be serialized just by directly copying all bits of the class. This is, in particular, the case for POD data types containing no pointer members, and which are neither versioned nor tracked. Some archives, such as non-portable binary archives can make us of this information to substantially speed up serialization.

为了表明按位序列化的可能性，使用了头文件 is_bitwise_serializable.hpp 中定义的类型特征:

To indicate the possibility of bitwise serialization the type trait defined in the header file is_bitwise_serializable.hpp is used:

这里是关键点:这个优化

Here are the key points: this optimization

• 在它适用的归档类型中是可选的

• 不适用于所有存档类型，例如

• is optional in the archive types where it does apply

• doesn't apply to all archive types e.g.

• 无法通过复制原始内存表示来实现需要可移植的二进制存档(因为它依赖于实现和平台)

• a binary archive that needs to be portable could not be implemented by copying out the raw memory representation (because it is implementation and platform depenedent)

文本存档可能不希望对此进行优化(例如，它有不同的目标，例如人类可读的 XML"，他们可能不想对您的 vector; 作为一个大的 bas64 编码 blob).

a text archive might not desire to optimize this (e.g. it has different goals to, like "human readable XML", they might not want to encode your vector<A> as a large bas64 encoded blob).

请注意，这也解释了 is_bit_wise_serializable<T> 并非部分专门用于具有 is_pod<T>::value == true 的任何类型(这在技术上可以很容易完成):

Note that this also explains that is_bit_wise_serializable<T> is not partially specialized for any type that has is_pod<T>::value == true (This could technically be done easily):

• 某些类可能对序列化它们的所有状态不感兴趣(因此，使用按位复制比仅选择有趣的位(双关语)要占用更多空间)

• some classes might not be interested in serializing all their state (so using bitwise copy would take a lot more space than just selecting the interesting bits (pun intended))

您没有特别问，但这是工作实现的样子:

You didn't ask, specifically, but this is what the working implementation would look like:

#include <boost/archive/binary_oarchive.hpp>
#include <boost/serialization/serialization.hpp>
#include <sstream>

struct A { int a; char b;
    template <typename Ar> void serialize(Ar& ar, unsigned) {
        ar & a;
        ar & b;
    }
};

BOOST_IS_BITWISE_SERIALIZABLE(A)

int main() {
    std::ostringstream oss;
    boost::archive::binary_oarchive oa(oss);

    A data { 1, 'z' };
    oa << data;
}

这篇关于提高序列化按位可序列化的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！