c++ - C++中客户端到服务器的数据传输问题

Question

我有一个使用 UDP 数据报将数组数据发送到服务器的客户端。为了能够跟踪我发送的数据，我用数字填充了我的数组。例如; 在客户端代码中我填buf[7] = 7;这个索引是为了表示分片总数。我希望在服务器中也能看到相同的。

但是，我的 buf in server 没有得到这些值。我必须在服务器中使用固定大小的字符数组 我无法更改它。 (我对数据进行了长时间的计算，因为与问题无关，所以没有放)

我要解释一下服务器结构:

• 我排长队。队列包含 3 个缓冲区。如果我获取数据的速度太快并且如果 buf 已满且其数据仍在文件中，则应使用 buf2等等。这是为了优化目的。
• 我使用一个线程，它将当前缓冲区数据写入一个文件并行的方式。我向线程发出信号，以便它知道何时开始写作。

我怀疑线程，当我评论它时，接收功能卡住了。我怀疑类型的数组，因为我发送了 char 指针数组但收到了固定大小的 char 数组。 (同样，我必须在接收时使用固定大小)

All I want is to see the data correctly transferred. When I look at server I want to see the same data at same indexes.

Here is my simple client code

void main(int argc, char* argv[]){
    WSADATA data;
    WORD version = MAKEWORD(2, 2);

    int wsOk = WSAStartup(version, &data);
    if (wsOk != 0)
    {
        cout << "Can't start Winsock! " << wsOk;
        return;
    }

    sockaddr_in server;
    server.sin_family = AF_INET; // AF_INET = IPv4 addresses
    server.sin_port = htons(...port no...); // Little to big endian conversion
    inet_pton(AF_INET, "...IP....", &server.sin_addr); // Convert from string to byte array

    SOCKET out = socket(AF_INET, SOCK_DGRAM, 0);
    int sendOk = 0;

    int *buf = new int[1450];

    //just fill it with ordered number to be able to trace from other side
    for (int i = 0; i < 1450; i++)
    {
        buf[i] = i;
    }

    buf[7] = 7; // this index represent TotalFragmentCount in UDP

    while (true) {

        for (int fragmentNumber = 1; fragmentNumber < 8; fragmentNumber++) {
            //I want to set this index to count fragment number when I look at server
            buf[9] = fragmentNumber;
            //I give it as char * array with reinterpret_cast, since sendto accepts pointer char array
            sendOk = sendto(out, reinterpret_cast<char*>(&buf), 1450, 0, (sockaddr*)& server, sizeof(server));
        }

    }

    if (sendOk == SOCKET_ERROR)
    {
        cout << "That didn't work! " << WSAGetLastError() << endl;
    }

    closesocket(out);
    WSACleanup();

}

Here is my server code

void writeToFile(char buf[], vector<uint16_t> &intData, ofstream &file) {


    while (true) {

        if (signaled == 1) {
            thread_mutex.lock();
            //writing...
            for (const auto& e : intData) {
                file << e << "\n";
            }
            buffers.pop(); // pops front which is written by this time
            buffers.push(buf);
            thread_mutex.unlock();
            break;
        }
        else{
            continue;
        }
    }
}
// Main entry point into the server
void main()
{
    WSADATA data;
    WORD version = MAKEWORD(2, 2);

    int wsOk = WSAStartup(version, &data);
    if (wsOk != 0)
    {
        // Not ok! Get out quickly
        cout << "Can't start Winsock! " << wsOk;
        return;
    }

    // Create a socket, notice that it is a user datagram socket (UDP)
    SOCKET in = socket(AF_INET, SOCK_DGRAM, IPPROTO_UDP);

    sockaddr_in serverHint;
    serverHint.sin_addr.S_un.S_addr = ADDR_ANY; // Us any IP address available on the machine
    serverHint.sin_family = AF_INET; // Address format is IPv4
    serverHint.sin_port = htons(4660); // Convert from little to big endian

    // Try and bind the socket to the IP and port
    if (bind(in, (sockaddr*)&serverHint, sizeof(serverHint)) == SOCKET_ERROR)
    {
        cout << "Can't bind socket! " << WSAGetLastError() << endl;
        return;
    }

    sockaddr_in client; // Use to hold the client information (port / ip address)
    int clientLength = sizeof(client); // The size of the client information


    char buf[1550] = { 0 }; //message gets here
    char buf2[1550]= { 0 };
    char buf3[1550]= { 0 };

    // Control buffer traffic with queue
    buffers.push(buf);
    buffers.push(buf2);
    buffers.push(buf3);

    std::thread thread_write(writeToFile, std::ref(buffers.front()), std::ref( intData), std::ref(file)); //front returns first element of the queue

    bytesIn = recvfrom(in, buffers.front(), 1550, 0, (sockaddr*)& client, &clientLength);
    cout << "";
    while (ReceivedFrameCount != 100)
    {
            ZeroMemory(&client, clientLength); // Clear the client structure

            // Wait for message
            bytesIn = recvfrom(in, buffers.front(), 1550, 0, (sockaddr*)& client, &clientLength);

            if (bytesIn == SOCKET_ERROR)
            {
                cout << "Error receiving from client " << WSAGetLastError() << endl;
                continue;
            }

            // Parse the byte array

            TotalFragmentCount = (uint16_t)(buf[6] << 8 | buf[7]);
            FrameFragmentNo = (uint16_t)(buf[8] << 8 | buf[9]);

            signaled = 1;


    }// end of while 


    // Close socket
    closesocket(in);

    file.close();
    WSACleanup();

    thread_write.join();
}

Answer 1

reinterpret_cast<char*>(&buf)不做你认为它做的事。应该是reinterpret_cast<char*>(buf) .和 1450应该是 1450*sizeof(int)因为你有 1450 个整数，每个整数都超过一个字节。

sendto 需要一个指向它将发送的数据的指针。如果你通过了&buf (指向 buf 的指针)那么您正在发送指针 buf .如果你通过了buf (指向缓冲区的指针)然后您正在发送缓冲区。

---

当你构建一个整数数组时，你的计算机将它们作为字节存储在内存中，如下所示:(也许 - 它是特定于平台的!)

Byte  0: 0 \
Byte  1: 0 |
Byte  2: 0 | int 0
Byte  3: 0 /
Byte  4: 1   \
Byte  5: 0   |
Byte  6: 0   | int 1
Byte  7: 0   /
Byte  8: 2     \
Byte  9: 0     |
Byte 10: 0     | int 2
Byte 11: 0     /

当服务器将这些视为字节时，它将在索引 8 处看到数字 2，依此类推。如果您希望它匹配，那么客户端也许应该创建一个字节数组(字符，与服务器相同)而不是整数。