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如何在C ++中使用Big Endian格式而不是Little Endian移位字节?

[英]How to shift the bytes using Big Endian format instead of Little Endian in C++?

I am writing Byte Array value into a file using Java with Big Endian Byte Order format.. Now I need to read that file from C++ program... 我正在使用Big Endian字节顺序格式的Java将字节数组值写入文件。现在,我需要从C ++程序读取该文件...

That Byte Array which I am writing into a file is made up of three Byte Arrays as described below- 我要写入文件的字节数组由三个字节数组组成,如下所述-

short employeeId = 32767;
long lastModifiedDate = "1379811105109L";
byte[] attributeValue = os.toByteArray();

I am writing employeeId , lastModifiedDate and attributeValue together into a single Byte Array and that resulting Byte Array I am writing into a file and then I will be having my C++ program which will retrieve that Byte Array data from file and then deserialize it to extract employeeId , lastModifiedDate and attributeValue from it. 我正在将employeeIdlastModifiedDateattributeValue一起写入一个字节数组,并将所得的字节数组写入文件中,然后将使用C ++程序,该程序将从文件中检索该字节数组数据,然后反序列化以提取employeeIdlastModifiedDateattributeValue

Below is my working Java code, which writes Byte Array value into a file with Big Endian format: 以下是我的工作Java代码,该代码将字节数组值写入Big Endian格式的文件中:

public class ByteBufferTest {

    public static void main(String[] args) {

        String text = "Byte Array Test For Big Endian";
        byte[] attributeValue = text.getBytes();

        long lastModifiedDate = 1289811105109L;
        short employeeId = 32767;

        int size = 2 + 8 + 4 + attributeValue.length; // short is 2 bytes, long 8 and int 4

        ByteBuffer bbuf = ByteBuffer.allocate(size); 
        bbuf.order(ByteOrder.BIG_ENDIAN);

        bbuf.putShort(employeeId);
        bbuf.putLong(lastModifiedDate);
        bbuf.putInt(attributeValue.length);
        bbuf.put(attributeValue);

        bbuf.rewind();

        // best approach is copy the internal buffer
        byte[] bytesToStore = new byte[size];
        bbuf.get(bytesToStore);

        writeFile(bytesToStore);

    }

    /**
     * Write the file in Java
     * @param byteArray
     */
    public static void writeFile(byte[] byteArray) {

        try{
            File file = new File("bytebuffertest");

            FileOutputStream output = new FileOutputStream(file);
            IOUtils.write(byteArray, output);           

        } catch (Exception ex) {
            ex.printStackTrace();
        }
    }
}

Now I need to retrieve Byte Array from that same file using the below C++ program and deserialize it to extract employeeId , lastModifiedDate and attributeValue from it. 现在,我需要使用以下C ++程序从同一文件中检索字节数组,并将其反序列化以从中提取employeeIdlastModifiedDateattributeValue I am not sure what is the best way on the C++ side. 我不确定C ++方面的最佳方法是什么。 Below is the code I have so far: 下面是我到目前为止的代码:

int main() {

    string line;

    std::ifstream myfile("bytebuffertest", std::ios::binary);

    if (myfile.is_open()) {

        uint16_t employeeId;
        uint64_t lastModifiedDate;
        uint32_t attributeLength;

        char buffer[8]; // sized for the biggest read we want to do

        // read two bytes (will be in the wrong order)
        myfile.read(buffer, 2);

        // swap the bytes
        std::swap(buffer[0], buffer[1]);

        // only now convert bytes to an integer
        employeeId = *reinterpret_cast<uint16_t*>(buffer);

        cout<< employeeId <<endl;

        // read eight bytes (will be in the wrong order)
        myfile.read(buffer, 8);

        // swap the bytes
        std::swap(buffer[0], buffer[7]);
        std::swap(buffer[1], buffer[6]);
        std::swap(buffer[2], buffer[5]);
        std::swap(buffer[3], buffer[4]);

        // only now convert bytes to an integer
        lastModifiedDate = *reinterpret_cast<uint64_t*>(buffer);

        cout<< lastModifiedDate <<endl;

        // read 4 bytes (will be in the wrong order)
        myfile.read(buffer, 4);

        // swap the bytes
        std::swap(buffer[0], buffer[3]);
        std::swap(buffer[1], buffer[2]);

        // only now convert bytes to an integer
        attributeLength = *reinterpret_cast<uint32_t*>(buffer);

        cout<< attributeLength <<endl;

        myfile.read(buffer, attributeLength);


        // now I am not sure how should I get the actual attribute value here?

        //close the stream:
        myfile.close();
    }

    else
        cout << "Unable to open file";

    return 0;
}

I have specifically set the storage Java-side to big-endian which means I know where each byte belongs I guess. 我专门将存储Java端设置为big-endian,这意味着我知道每个字节都在哪里。 So how do I code it while shifting the bytes into the proper position for each and every value? 那么,如何在将字节移动到每个值的正确位置时进行编码? Right now I am coding it as little-endian I guess which is not what I want... 现在我正在将其编码为小端,我想这不是我想要的...

I was reading somewhere that I can use ntoh in C++ to deserialize the Byte Array.. Not sure whether htons will be much better solution as compared to what I have currently?.. 我在某处读书,可以在C ++中使用ntoh对字节数组进行反序列化。.不确定htons是否比我目前拥有的解决方案好得多?

If yes, then I am not sure how to use that in my current C++ code? 如果是,那么我不确定如何在当前的C ++代码中使用它?

Can anybody take a look on C++ code and see what I can do to improve it, as I don't think it is looking much efficient? 谁能看看C ++代码,看看我能做些什么来改进它,因为我认为它看起来效率不高? Any better way to deserialize the Byte Array and extract relevant information on the C++ side? 还有什么更好的方法可以在C ++方面反序列化字节数组并提取相关信息?

If the Java and the C++ Code is developed by you then it's perhaps better to use the Google Protocol Buffer ( https://developers.google.com/protocol-buffers/docs/overview ) instead to write your own serializer / deserializer. 如果您是由Java和C ++代码开发的,那么最好使用Google协议缓冲区( https://developers.google.com/protocol-buffers/docs/overview )来编写自己的序列化器/反序列化器。

If you really want to write your own implementation the best way is to write a buffer class which receives the byte-stream as parameter (eg as Constructor parameter) and make some access methodes readShort / readLong / readInt / readByte ... and only swap the needed bytes. 如果您真的想编写自己的实现,最好的方法是编写一个缓冲区类,该缓冲区类将字节流作为参数(例如,构造函数参数)接收,并使某些访问方法readShort / readLong / readInt / readByte ...仅交换所需的字节。

class ByteBuffer{
  explicit ByteBuffer(uint8_t* byteStream, uint16_t streamLength);
  uint8_t readUInt8(uint16_t readPos)const {return m_byteStream[readPos];} // no conversion needed
  uint16_t readUInt16(uint16_t readPos)const {
    const uint8_t byteCount = 2;
    union{
      uint16_t u16;
      uint8_t u8[byteCount];
    }tmp;
    for(uint8_t i=0; i<byteCount; ++i){
      tmp.u8[i] = readUInt8(readPos+i*8);
    }
    return ntohs(tmp.u16); // do conversion
  }
  ...
}

Checks for read behind the buffer is missing here. 此处缺少对缓冲区后面的读取的检查。 If your code should be portable then you have use ntohl / ntohs (see: http://forums.codeguru.com/showthread.php?298741-C-General-What-do-ntohl%28%29-and-htonl%28%29-actually-do ). 如果您的代码应该是可移植的,那么您可以使用ntohl / ntohs(请参阅: http ://forums.codeguru.com/showthread.php?298741-C-General-What-do-ntohl%28%29-and-htonl% 28%29-actually-do )。 If you swap the bytes by our own then your code is not portable (runs only on Little-Endian machines). 如果您通过我们自己交换字节,那么您的代码将不可移植(仅在Little-Endian计算机上运行)。 If you use ntoh then it would also run on a such machine. 如果您使用ntoh,那么它也可以在这样的机器上运行。

For convenience I would also write a wrapper class where you can read and write your fields (eg employeeId) directly: 为了方便起见,我还将编写一个包装器类,您可以在其中直接读写字段(例如,employeeId):

class MyBuffer{
  uint16_t readEmployeeId()const{return m_Buffer.readuint16(EmployeeId_Pos);}
  ....
  static const uint16_t EmployeeId_Pos = 0;
  ....
}

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