基於匹配字符串模式來簡化函數

Question

問題：我是C++的新手，在編寫以下代碼後，似乎應該有一種方法來縮短它。也許通過某種方式匹配字符串？這將如何完成？

該函數接收通過串口接收到的字符串message，並根據message設置pinValues[]數組中某個元素的值。將設置的值取決於\n之前的最後一個字符H或L。

字符串模式：(a number)(H or L)\n

例如：message == "4H\n"將所述第五元件pinValues[4]設置爲HIGH。字符串開頭的數字可以是1至2位數字。

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

我會對字符串進行一些理智的檢查，同時從前兩個字符中提取鍵和值。如果您不需要合法性檢查的消息，也可能是短

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

雖然你可能想使一個長一點，即檢查字符串長度，並且該2個字符的檢查是在正確的範圍內。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; 
    } 
} 

編輯：您還可以擴展，要檢查兩個主要數字，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

假設String實際上是一個std::string或具有相同的接口，同時還假設一個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

這可能不是官方的「C++」 - 批准這樣做的方式，但你可以這樣做：

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;