在 C++ 中寻找 MemoryStream

发布于 2024-09-06 05:13:43 字数 429 浏览 3 评论 0原文

在 C# 的奇妙世界中，我可以创建一个内存流而不指定其大小，写入其中，然后只获取底层缓冲区。

我怎样才能在 C++ 中做同样的事情？基本上我需要做的是：

memory_stream  ms(GROW_AS_MUCH_AS_YOU_LIKE);

ms << someLargeObjects << someSmallObjects << someObjectsWhosSizeIDontKnow;

unsigned char* buffer = ms.GetBuffer();
int bufferSize = ms.GetBufferSize();

rawNetworkSocket.Send(buffer, bufferSize);

顺便说一句，我在我的项目中得到了提升，尽管我不太熟悉它。

谢谢。

原文

In the wonderful world of C# i can create a memory stream without specifying its size,
write into it and then just take the underlying buffer.

How can i do the same in c++? basicly i need to do:

memory_stream  ms(GROW_AS_MUCH_AS_YOU_LIKE);

ms << someLargeObjects << someSmallObjects << someObjectsWhosSizeIDontKnow;

unsigned char* buffer = ms.GetBuffer();
int bufferSize = ms.GetBufferSize();

rawNetworkSocket.Send(buffer, bufferSize);

By the way I have boost in my project though I'm not all that familiar with it.

Thank you.

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无悔心 2024-09-13 05:13:43

#include <sstream>

std::ostringstream  buffer; // no growth specification necessary
buffer << "a char buffer" << customObject << someOtherObject;

std::string contents = buffer.str();
size_t bufferSize = contents.size();

rawNetworkSocket.Send(contents); // you can take the size in Send

使用这种方法，您必须在收到结果的地方解析结果（因为上面的代码只是将您的数据转换为非结构化字符串。

它的另一个问题是，由于 C++ 不支持反射，您必须定义运算符 < Custom 类的代码：

template<typename C, typename T>
std::basic_ostream<C,T>& operator << (
    std::basic_ostream<C,T>& out, const Custom& object)
{
    out << object.member1 << "," << object.member2 /* ... */ << object.memberN;
    return out;
}

如果您想要结构化序列化，请查看 boost::序列化。

#include <sstream>

std::ostringstream  buffer; // no growth specification necessary
buffer << "a char buffer" << customObject << someOtherObject;

std::string contents = buffer.str();
size_t bufferSize = contents.size();

rawNetworkSocket.Send(contents); // you can take the size in Send

Using this approach you will have to parse the result where you receive it (as the code above just transforms your data into an unstructured string.

Another problem with it is that since C++ doesn't support reflection, you will have to define operator << for your objects. This is the code for a Custom class:

template<typename C, typename T>
std::basic_ostream<C,T>& operator << (
    std::basic_ostream<C,T>& out, const Custom& object)
{
    out << object.member1 << "," << object.member2 /* ... */ << object.memberN;
    return out;
}

If you want structured serialization, have a look at boost::serialization.

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漫雪独思 2024-09-13 05:13:43

为此，您可能需要查看 std::stringstream 。该流将根据需要增长。除非您想将对象保留为二进制而不是 ASCII，在这种情况下您可以查看 Streambuf 对象和实现。

请注意，C++ 没有反射或双重/多重分派，因此您必须自己为未知大小的对象提供支持：

class unknown_base {
   virtual void dump( std::ostream & ) const;
};
std::ostream& operator<<( std::ostream& o, unknown_base const & obj ) {
   obj.dump( o );
   return o;
}
std::string serialize( std::vector<unknown_base*> const & data ) {
   std::ostringstream st;
   for ( std::vector<unknown_base*>::const_iterator it = data.begin(), end = data.end();
         it != end; ++it ) {
      st << **it; // double dereference: iterator, pointer
   }
   return st.str();
}

You may want to look at std::stringstream for that purpose. The stream will grow as required. Unless you want to leave the objects in binary instead of ASCII, in which case you could take a look at the streambuf objects and implementations.

Note that C++ does not have reflection or double/multiple dispatch, so you will have to provide support for the unknown sized object yourself:

class unknown_base {
   virtual void dump( std::ostream & ) const;
};
std::ostream& operator<<( std::ostream& o, unknown_base const & obj ) {
   obj.dump( o );
   return o;
}
std::string serialize( std::vector<unknown_base*> const & data ) {
   std::ostringstream st;
   for ( std::vector<unknown_base*>::const_iterator it = data.begin(), end = data.end();
         it != end; ++it ) {
      st << **it; // double dereference: iterator, pointer
   }
   return st.str();
}

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