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我正在使用 dsPic33 尝试接收一个 11 字节的字符串并将其放入一个数组中，但没有成功地完全接收到它。我发送的字符串是“$123456789#”，应该由图片接收。我试过使用下面的代码。任何帮助将不胜感激。

char inBytes[11];
int i;
unsigned char temp;

while (U1STAbits.URXDA != 0)
{
  temp = U1RXREG;
  if (temp == '$')
  {
      inBytes[0] = temp;
      for(i = 1; i < 10; i++)
      {
        if (U1STAbits.URXDA != 0)
        inChar = U1RXREG;
        inBytes[i] = inChar;
      }
  }
4

2 回答 2

1

for(i = 1; i < 10; i++)从索引 1 开始保存数据并在 9 处停止，只有 9 个字节。更改< 10为<= 10或< 11。

于 2015-01-09T14:07:45.760 回答
1

jolati 有一个很好的观点，即最终值太低而无法获得 11 个字节，但我必须补充一点，您必须等待其他字节可用，然后才能读取它们。

在你的例子中；

char inBytes[11];
int i;
unsigned char temp;

while (!U1STAbits.URXDA ); //Wait until at least one byte is available

temp = U1RXREG;
if (temp == '$')
{
    inBytes[0] = temp;

    for(i = 1; i < 11; i++) //Loop over range i = 1 to 10 inclusively
    {
        while (!U1STAbits.URXDA ); //Wait until at least one byte is available
        inBytes[i] = U1RXREG;
    }
}

理想情况下，您可以使用中断以非阻塞方式执行此操作，以便在数据进入时处理数据，但是，如果您不能使用中断，则始终可以使用非阻塞轮询，例如：

void AsyncRX()
{
    //Note that the static variables keeps their value between invocations of the
    //  function. i is set to 0 only on the first run of this function, it keeps 
    //  its value on every other run after that.
    static int i = 0;
    static char inBytes[11];

    //Nothing more to do until there is at least 1 byte available
    if( !U1STAbits.URXDA ) 
        return;

    //Save the byte and test that our message starts with $
    inBytes[i] = U1RXREG;
    if( inBytes[0] != '$' )
        return;

    //Test the counter to see if we have a full 11 bytes
    i++;
    if( i < 11 )
        return;

    //Do something with your 11 bytes
    //...
    //...

    //Reset the counter for the next message
    i = 0;
}

对于中断示例，您可以简单地获取轮询版本并将其放入 ISR。下面是一个例子。请注意，我不知道您使用的是哪个 dsp33，而且我有一段时间没有在高端内核（带有向量表）中编程中断，因此您可能需要进行一两次更改。另请注意，您需要通过设置适当的寄存器来启用中断，因为默认情况下不启用它们。

void __attribute__ ((interrupt, no_auto_psv)) _U1RXInterrupt(void) 
{
    //Note that the static variables keeps their value between invocations of the
    //  function. i is set to 0 only on the first run of this function, it keeps 
    //  its value on every other run after that.
    static int i = 0;
    static char inBytes[11];

    //Reset the interrupt flag
    IFS0bits.U1RXIF = 0;

    //Use up all bytes in the buffer (the interrupt can be set to only fire 
    //  once the buffer has multiple bytes in it).
    while( U1STAbits.URXDA ) 
    {
        //Save the byte and test that our message starts with $
        inBytes[i] = U1RXREG;
        if( inBytes[0] != '$' )
            continue;

        //Test the counter to see if we have a full 11 bytes
        i++;
        if( i < 11 )
            continue;

        //Do something with your 11 bytes
        //...
        //...

        //Reset the counter for the next message
        i = 0;
    }
}
于 2015-01-10T03:08:16.077 回答