如何将二进制数据复制到字符串流

Question

I have a std::vector<int> and I want serialize it. For this purpose I am trying to use a std::stringstream

 vector<int> v;
 v.resize(10);
 for (int i=0;i<10;i++)
 v[i]=i;


 stringstream ss (stringstream::in | stringstream::out |stringstream::binary);

However when I copy the vector to the stringstream this copy it as character

ostream_iterator<int> it(ss);
copy(v.begin(),v.end(),it);

the value that inserted to buffer(_Strbuf) is "123456789"

I sucssesed to write a workaround solution

for (int i=1;i<10;i++)
   ss.write((char*)&p[i],sizeof(int));

I want to do it something like first way by using std function like copy

thanks Herzl

Answer 1

Actually, this is your workaround but it may be used with std::copy() algorithm.

   template<class T>
   struct serialize
   {
      serialize(const T & i_value) : value(i_value) {}
      T value;
   };

   template<class T>
   ostream& operator <<(ostream &os, const serialize<T> & obj)
   {
      os.write((char*)&obj.value,sizeof(T));
      return os;
   }

Usage

ostream_iterator<serialize<int> > it(ss);
copy(v.begin(),v.end(),it);

Answer 2

I know this is not an answer to your problem, but if you are not limited to the STL you could try ( boost serialization ) or google protocol buffers

Boost even has build-in support for de-/serializing STL containers (http://www.boost.org/doc/libs/1_45_0/libs/serialization/doc/tutorial.html#stl)

Answer 3

To use std::copy with ostream::write, you'd need to write your own output iterator that knows how to correctly serialize the type. That said, I'm not sure what you expect to gain from this approach, but here's a first pass at that for an example:

struct ostream_write_int
  : std::iterator<std::output_iterator_tag, int, void, void, void>
{
  std::ostream *s;
  ostream_write_int(std::ostream &s) : s (&s) {}

  ostream_write_int& operator++() { return *this; }
  ostream_write_int& operator++(int) { return *this; }
  ostream_write_int& operator*() { return *this; }

  void operator=(int x) {
    s->write(reinterpret_cast<char*>(&x), sizeof(x));
  }
};

This could be templated only if you defer the serialization logic to some other function (as the formatted stream iterators do to operator<<).

Answer 4

像弗雷德一样，我没有看到这一点，你有效地尝试做的是：

ss.rdbuf()->sputn(reinterpret_cast<char*>(&v[0]), sizeof(int) * v.size());