Question

问题：我是C ++的新手，在写完下面的代码之后，似乎应该有一种缩短它的方法。也许以某种方式匹配字符串？怎么会这样做？

该函数接收通过串行端口接收的字符串message，并根据pinValues[]设置message数组的特定元素的值。将设置的值由H之前的最后一个字符L或\n确定。

字符串模式：(a number)(H or L)\n

例如：message == "4H\n"会将第5个元素pinValues[4]设置为HIGH。字符串开头的数字可以是 1到2位。

void setPinValues(String message) {
     if( message == "1H\n" ) {
          pinValues[1] = HIGH;
      }
      if( message == "1L\n" ) {
          pinValues[1] = LOW;
      }
      if( message == "2H\n" ) {
          pinValues[2] = HIGH;
      }
      if( message == "2L\n" ) {
          pinValues[2] = LOW;
      }
      if( message == "3H\n" ) {
          pinValues[3] = HIGH;
      }
      if( message == "3L\n" ) {
          pinValues[3] = LOW;
      }
      if( message == "4H\n" ) {
          pinValues[4] = HIGH;
      }
      if( message == "4L\n" ) {
          pinValues[4] = LOW;
      }
      if( message == "5H\n" ) {
          pinValues[5] = HIGH;
      }
      if( message == "5L\n" ) {
          pinValues[5] = LOW;
      }
      if( message == "6H\n" ) {
          pinValues[6] = HIGH;
      }
      if( message == "6L\n" ) {
          pinValues[6] = LOW;
      }
}

Answer 1

我会在从前两个字符中提取键和值时对字符串进行一些完整性检查。如果您不需要完整性检查消息，它可能就像

一样短

void setPinValues(String message) {
    pinValues[ message[0] - '0' ] = (message[1] == 'H') ? HIGH:LOW;
}

虽然您可能想要更长一点，即检查字符串长度，并检查2个字符是否在正确的范围内。即

void setPinValues(string message) {

  if ( 
      message.size() >= 2 
      and 
      message[0] >= '1' and message[0] <= '6' 
      and (message[1]=='H' or message[1]=='L')
       ) {
    pinValues[ message[0] - '0' ] = (message[1] == 'H') ? HIGH:LOW;
  }
}

编辑：您也可以将其扩展到检查两个前导数字，即

int n, off=0;


if ( s[off] <= '9' and s[off] >= '0') 
{
  n = s[off++] - '0';
}
if ( s[off] <= '9' and s[off] >= '0')  
{
  n = 10*n + s[off++] - '0';
}
if (off > 0 and (s[1]=='H' or s[1]=='L')) {
  pinValues[ message[0] - '0' ] = (message[1] == 'H') ? HIGH:LOW;
}

Answer 2

假设String实际上是std::string或具有相同的接口，并且还假设与ASCII兼容的字符集...

void setPinValues(String message) {
    const size_t sz = message.size();

    // input validation, ignore the message if it doesn't fit the pattern
    // you can remove this "if" block if the message has already been validated
    if (   (sz < 3) || (sz > 4)
           // note how message[0] will be checked twice if sz == 3
           // once as message[0] and once as message[sz -3]
           // but if sz == 4 we check message[0] and message[1]
        || (message[0] < '0') || (message[0] > '9')
        || (message[sz - 3] < '0') || (message[sz - 3] > '9')
        || ((message[sz - 2] != 'H') && (message[sz - 2] != 'L'))
        || (message[sz - 1] != '\n'))
      return;

    // convert the first or two characters to a number
    int pinNumber = message[0] - '0';
    if (sz == 4)
        pinNumber = (pinNumber * 10) + (message[1] - '0');

    // additional check to verify the pin number is in the correct range
    if ((pinNumber < 1) || (pinNumber > 6))
        return;

    // apply
    pinValues[pinNumber] = (message[sz - 2] == 'H' ? HIGH : LOW);
}

Answer 3

这可能不是官方的“C ++”批准的方式，但你可以这样做：

unsigned int pinNo = 0;
unsigned char level = 0;
int result = sscanf(message.c_str(), "%u%c", &pinNo, &level);
if (result < 2)
  // it failed
if (pinNo > 6)
  // bad data
levelVal = (level == 'H') ? HIGH : LOW;

基于匹配字符串模式简化函数

3 个答案: