0

我从 UART 获取字符串数据并将其转换为十六进制。转换为 hex 后，我有两个唯一的分隔符“06”和​​“07”。我想获取这两个分隔符之间的数据。我试图根据我所说的构建下面的代码。但是在收到 6-7 个输出后，我得到了分段错误错误。

 std::string sub_between(std::string start,std::string stop,std::string input)
    {
       auto start_index = input.find(start);
       if (start_index != std::string::npos) 
       {
          auto stop_index = input.find(stop,start_index+start.size());
          if (stop_index != std::string::npos)
          {
            return {input.begin() + start_index+start.size(), input.begin() + stop_index};
          }
       }
        //  return empty string
        return {};
    }

    int main(int argc, char ** argv) 

{
  int uart0_filestream = -1;
  int uart2_filestream = -1;
  unsigned char rx_buffer[1024];
  memset(&rx_buffer[0], 0, 1023);

  unsigned char tx_buffer[1024];
  memset(&tx_buffer[0], 0, 1023);

  std::stringstream ss;


  if(uart0_filestream == -1)
  {
    uart0_filestream = open("/dev/ttyS0", O_RDWR | O_NOCTTY | O_NDELAY);
    //uart0_filestream = open("/dev/ttyS0", O_NOCTTY );
    
    if(uart0_filestream == -1)
    {
        std::cout << "UARTListener::Start Unable to open UART" << std::endl;
    }
    else
    {
      struct termios options;
      tcgetattr(uart0_filestream, &options);
      options.c_cflag = B115200 | CS8 | CLOCAL | CREAD;   //<Set baud rate
      options.c_iflag = IGNPAR;
      options.c_oflag = 0;
      options.c_lflag = 0;
      //options.c_line = '\n';
      //options.c_cc[VEOL] = 0; 
      tcflush(uart0_filestream, TCIFLUSH);
      tcsetattr(uart0_filestream, TCSANOW, &options);
    }
  }

    while(true)
    {
      memset(&rx_buffer[0], 0, 1024);
      int rx_length = read(uart0_filestream, (void*) rx_buffer, 1023);
      if(rx_length > 0)
      {
        rx_buffer[rx_length] = '\0';
        //std::string txt = std::string((char*) rx_buffer);

        for (int i = 0; i < rx_length ; i++)
        {
           ss << std::setfill('0') << std::setw(2)<< std::hex << (int)rx_buffer[i];
           //ss << std::hex << (int)rx_buffer[i];
        }
        std::string startDEL = "06";
        std::string stopDEL =  "07";

        std::string concat = std::accumulate(std::begin(ss.str()),std::end(ss.str()),std::string(""));
        std::cout << "Pulled Data : " << ss.str() << std::endl;
        std::cout << "String : " << sub_between(startDEL,stopDEL,concat) << std::endl;

      //std::cout << txt << std::endl;
      }
    }
    return 0;
  }

- 输出 - 

    Pulled Data : 42
    String : 
    Pulled Data : 426561636f6e
    String : 
    Pulled Data : 426561636f6e283229
    String : 
    Pulled Data : 426561636f6e28322920466f756e
    String : 
    Pulled Data : 426561636f6e28322920466f756e645b355d
    String : 
    Segmentation fault
4

0 回答 0