为什么C ++ msgpack-c在数字10(0x0A)之前添加数字13(0x0D),而C#MessagePack-CSharp却没有?

时间:2019-05-13 15:20:53

标签: c# c++ msgpack binary-serialization

我试图使用MessagePack在C ++和Windows的C#中将0到127的整数序列化,但是结果不一样。 msgpack-c在0x09和0x0A之间插入0x0D,但MessagePack-CSharp不插入。为什么会这样?

操作系统:Windows 10

IDE:Visual Studio 2019

C#

图书馆:

https://github.com/neuecc/MessagePack-CSharp

代码:

using System.Collections.Generic;
using System.IO;

class Program
{
    static void Main(string[] args)
    {
        using (FileStream fileStream = new FileStream("CSharp.msgpack", FileMode.Create, FileAccess.Write))
        {
            List<int> list = new List<int>();

            for (int i = 0; i < 128; ++i)
            {
                list.Add(i);
            }

            MessagePackSerializer.Serialize(fileStream, list);
        }
    }
}

结果:

CSharp

C ++

图书馆:

https://github.com/msgpack/msgpack-c

代码:

#include <msgpack.hpp>
#include <vector>
#include <iostream>
#include <fstream>

int main(void)
{
    std::ofstream OutputFileStream;

    std::vector<int> list;

    for (int i = 0; i < 128; ++i)
    {
        list.push_back(i);
    }

    OutputFileStream.open("Cpp.msgpack");

    msgpack::pack(OutputFileStream, list);

    OutputFileStream.close();
}

结果:

Cpp

1 个答案:

答案 0 :(得分:5)

由于您以文本模式在c ++中打开文件,因此如果Windows中不存在,每个\n(ASCII 10)都将以\r(ASCII 13)作为前缀。您需要以二进制模式打开文件,以免发生这种情况。

OutputFileStream.open("Cpp.msgpack", std::ofstream::binary);