通过sendto发送UDP数据包

问题描述

我想通过 sendto 发送一个带有 4 字节地址、1 个短整数和 4 字节整数距离的数据报 UDP 消息。所以它应该正好是 9 字节。我该怎么做？我已经尝试转换这些值并将它们放入缓冲区，但没有成功。

sendto(sockfd,&message,sizeof(message),(struct sockaddr*) &server_address,sizeof(server_address));

编辑： 我也尝试创建，但它有 12 个字节。

 struct msg {
      uint32_t dist;
      uint8_t num ;
       struct in_addr addr;
  
};

解决方法

最便携的方法是自己将字节组装到缓冲区中：

uint8_t buf[64];  // make sure this is large enough to hold the bytes
uint8_t * b = buf;

uint32_t neDist = htonl(msg.dist);   // send multi-byte values in big-endian form
memcpy(b,&neDist,sizeof(neDist));
b += sizeof(neDist);

memcpy(b,&msg.num,sizeof(msg.num));
b += sizeof(msg.num);

uint32_t neAddr = htonl(msg.addr.s_addr);
memcpy(b,&neAddr,sizeof(neAddr));
b += sizeof(neAddr);

uint32_t numBytesToSend = (b-buf);
sendto(sockfd,buf,numBytesToSend,(struct sockaddr*) &server_address,sizeof(server_address));

...然后在接收方，例如：

uint8_t buf[64];  // make sure this is large enough to hold the bytes
int32_t numBytesReceived = recvfrom(sockfd,sizeof(buf),...);
if (numBytesReceived > 0)
{
    struct msg m;

    uint8_t * b = buf;
    uint8_t * fib = buf+numBytesReceived;  // pointer to the first invalid byte

    if ((fib-b) >= sizeof(m.dist))
    {
       uint32_t neDist;
       memcpy(&neDist,b,sizeof(neDist));
       b += sizeof(neDist);
       m.dist = ntohl(neDist);  // convert from big-endian back to local-endian

       if ((fib-b) >= sizeof(m.num))
       {
          memcpy(&m.num,sizeof(m.num));
          b += sizeof(m.num);

          if ((fib-b) >= sizeof(m.addr.s_addr))
          {
             uint32 neAddr;
             memcpy(&neAddr,sizeof(neAddr));
             m.addr.s_addr = ntohl(neAddr);

             printf("At this point,(m) is fully populated and ready for use\n");
          }
       }
    }
}