6

我把事情复杂化了吗？

我正在构建我的代码以通过 UART 从 8051 微型设备与外围设备通信。外设响应来自主机的命令，并且一次只能响应一个命令。这是一个简单的发送和接收协议。(tx1, rx1, tx2, rx2, tx3, rx3) 每个 TX 消息以 CR 结束，每个响应以 > 结束。在收到对最后一条消息的回复之前，我无法发送新消息。如果我启用该选项，响应也可以在开头回显原始 TX 消息（但这会导致更多流量）

一个示例消息是：

  • TX：你好
  • RX：世界！>

或使用回声选项...

  • TX：你好
  • RX：你好\r世界！>

选项 A 诸如 getHello 之类的函数将包含发送和接收。并行 ISR 例程将收集传入字节并在接收到“>”字符时抛出一个标志。

char* getHello(char * buf){
    sendMsg("Hello\r");
    delay(10ms); //wait a little bit

    //wait for receive to come in or timeout to occur
    while(!receiveFlag || !timeoutFlag);  //thrown by ISR
    receiveMsg(buf);
    //parse the message and do some other stuff
    return buf;
}

优点：

  • 一切都包含在一个功能中。
  • 更容易调试

缺点：

  • 这个函数是阻塞的，如果外设从不响应，可能会挂起，因此必须实现超时。
  • 消息不能乱序接收，必须串联（即tx1、rx1、tx2、rx2、tx3、rx3）

选项 B 采用并行方法。将创建两个单独的函数。一种用于发送消息，另一种用于在收到 ISR 的响应时进行顶点处理。

void sendHello(){
    sendMsg("Hello\r");
    //do some other stuff if needed
}

char* receiveMsg(char * buf){
    //figure out from echo print what the tx message was
    //use a switch statement to decide which response parser to call
    switch(txMessage){ //pseudo code
    case "Hello":
        receiveMsg(buf);
        //parse the message and do some other stuff
        break;
    }
    return buf;
}

优点：

  • 可以处理乱序返回的并行消息，因为它依赖于 tx 消息的回显打印来确定如何解析它。（即，tx1、tx2、tx3、rx1、rx2、rx3）

缺点：

  • 调试相当困难
  • 产生多个线程
  • 很多额外的代码
  • 不值得，因为消息肯定会按顺序返回

现在，我正在做选项 B，但随着我继续这个项目，我开始觉得这变得过于复杂了。我很好奇你们是怎么想的。

谢谢！

4

2 回答 2

4

不过，我倾向于做这种事情，我倾向于有一个单独的串行端口“类”（ struct + functions ）和一个位于串行端口顶部的协议类。我一直在我的嵌入式系统中使用这些。这为您提供了两全其美，阻塞同步调用和异步调用，因此您可以伪多任务。

typedef struct serial_port_s serial_port;
typedef void (*serial_on_recived_proc)(serial_port* p);
typedef struct serial_port_s{
    bool timeoutFlag;
    bool receiveFlag;
    void* context;
    serial_on_recived_proc response_handler;
};

void send_serial(serial_port* p, char* message)
{
    //SendMsg?
}
void receive_serial(serial_port* p, char* response)
{
    //receiveMsg?
}

bool has_data(serial_port* p)
{
    return p->receiveFlag;
}

bool has_timed_out(serial_port* p)
{
    return p->timeoutFlag;
}
bool is_serial_finished(serial_port* p)
{
    return has_data(p) || has_timed_out(p); 
}

bool serial_check(serial_port* p)
{
    if(is_serial_finished(p) && p->response_handler != NULL)
    {
       p->response_handler(p)
       p-> response_handler = NULL;
       return true;
    }
    return false;
}

void send(serial_port* p, char* message, char* response)
{
    p->response_handler=NULL;
    send_serial(p, message);
    while(!is_serial_finished(p));
    receive_serial(p, response);
}

void sendAsync(serial_port* p, char* message, serial_on_recived_proc handler, void* context)
{
    p->response_handler = handler;
    p->context = context;
    send_serial(p, message);
}

void pow_response(serial_port* p)
{
    // could pass a pointer to a struct, or anything depending on what you want to do
    char* r = (char*)p->context;  
    receive_serial(p, r);
    // do stuff with the pow response
}

typedef struct
{
   char text[100];       
   int x;
   bool has_result;
} bang_t;

void bang_parse(bang_t* bang)
{
   bang->x = atoi(bang->text);
}

void bang_response(serial_port* p)
{
    bang_t* bang = (bang_t*)p->context;  
    receive_serial(p, bang->text);
    bang_parse(bang);
    bang->has_result=true;
}

void myFunc();
{
    char response[100];
    char pow[100];
    bang_t bang1;
    bang_t bang2;
    serial_port p; //
    int state = 1;
    // whatever you need to do to set the serial port

    // sends and blocks till a response/timeout
    send(&p, "Hello", response);
    // do what you like with the response

    // alternately, lets do an async send...
    sendAsync(&p, "Pow", pow_response, pow);       

    while(true)
    {
        // non block check, will process the response when it arrives               
        if(serial_check(p))
            {
              // it has responded to something, we can send something else...

              // using a very simple state machine, work out what to send next.
              // in practice I'd use enum for states, and functions for managing state
              // transitions, but for this example I'm just using an int which
              // I just increment to move to the next state
              switch(state)
              {
              case 1: 
                 // bang1 is the context, and will receive the data
                 sendAsync(&p, "Bang1", bang_response, &bang1);
                 state++; 
                 break;
              case 2:
                 // now bang2 is the context and will get the data...
                 sendAsync(&p, "Bang2", bang_response, &bang2);
                 state++; 
                 break;
              default:
                 //nothing more to send....
                 break;
              }
            }
        // do other stuff you want to do in parallel
    }
};
于 2012-11-23T03:31:35.097 回答
0

把事情简单化。ISR 例程必须非常快，所以对我来说最好的方法是拥有一个像这样的全局 RXBuffer：

#include <cstdint>
#include <deque>
#include <algorithm>

class RXBuffer {
public:
    friend class ISR;

    typedef std::deque<uint8_t>::const_iterator const_iterator;

    RXBuffer();
    size_t size() const { m_buffer.size(); }

    // read from the buffer in a container in the range [first, last)
    template <typename Iterator>
    void read(Iterator first, Iterator last, Iterator to)
    {
        // how many bytes do you want to read?
        size_t bytes_to_read = std::distance(first, last);

        if (bytes_to_read >= size())
        {
            // read the whole buffer
            std::copy(begin(), end(), first);

            // empty the buffer
            m_buffer.clear();

            return size();
        }
        else
        {
            // copy the data
            copy(begin(), begin() + bytes_to_read, firt);

            // now enque the element
            m_buffer.erase(begin(), begon() + bytes_to_read);

            return bytes_to_read;
        }
    }

private:
    void put(uint8_t data)
    {
        // check buffer overflow
        m_buffer.push_back(data);
    }

    const_iterator begin() const { return m_buffer.begin(); }
    const_iterator end() const { return m_buffer.end(); }

private:
    std::deque<uint8_t> m_buffer;           // buffer where store data
    size_t m_size;                          // effective size of the container
};

class ISR {
public:
    ISR(RXBuffer& buffer) : m_buffer(buffer) {}

    // ISR Routine
    void operator () (uint8_t data)
    {
        m_buffer.put(data);
    } 

private:
    RXBuffer& m_buffer;
};
RXBuffer g_uart1_rx_buffer;

现在你有了 ISR 和一个 RXBuffer 来搜索数据，所以你需要一些东西来包装 UART 函数。您可以按如下方式实现：

class UART {
public:
    UART(unsigned int uart_device, RXBuffer& rx_buffer) :
        m_uart(uart_device), m_buffer(rx_buffer)
    {
    }

    unsigned int uart_device() const { return m_uart; }

    // set the timeout during a read operation
    void timeout(unsigned ms) { m_timer.countdown(ms); }

    template <typename InputIterator>
    void read(InputIterator first, InputIterator last)
    {
        // start the timer
        m_timer.start();

        size_t size = std::distance(first, last);
        size_t read_bytes = 0;

        while (read_bytes != size && !m_timer.is_expired())
        {
            read_bytes += m_buffer.read(first + read_bytes, last);
        }

        if (read_bytes != size) throw std::exception("timeout");
    }

    template <typename OutputIterator>
    void send(OutputIterator first, OutputIterator last)
    {
        size_t size = std::distance(first, last);
        uart_send(m_uart, &(*first), size);
    }

private:
    unsigned int m_uart;
    RXBuffer& m_buffer;
    timer m_timer;
};
于 2012-11-23T11:46:04.697 回答