C++如何重新啓動if語句？

Question

說你的代碼是：

cout << "You wake up in a room. There is a small lit candle and a door. What do you do?" << endl; 
     string startout; 
     getline(cin, startout); 
      if (startout == "Door" || startout == "DOOR" || startout == "door" || startout == "Open Door" || startout == "open door" || startout == "OPEN DOOR" || startout == "Open door" || startout == "Open the door" || startout == "open the door") { 
       cout << " " << endl; 
       cout << "You move to the door but its too dark to see anything besides the outline of it." << endl; 
      } else if (startout == "Candle" || startout == "CANDLE" || startout == "candle" || startout == "Pick Up Candle" || startout == "PICK UP CANDLE" || startout == "pick up candle" || startout == "Pick up candle") { 
       cout << "You pick up the candle then move to the door. With the light from the candle you can see the door well. What do you do?" << endl; 
      }; 

如果使用的代碼類型的東西，除了「門」或「蠟燭」（或任何變體），還可以將代碼是什麼重啓if語句，因此重新誰是 - 問第一個問題？

例如：用戶輸入：「跳舞」

輸出：「我不明白，‘跳舞’。 你是做什麼？」

或類似的東西。

Answer 1

您可以在外部添加一個while(true)循環，並添加if條件，該條件在正確輸入時會中斷。

另外，函數getline命令之後，如果別的conditon之後添加其他條件：

else 
    cout << "I don't understand " << startout << ". What do you do?" << endl; 

在其他if和else if子句，添加一個break;命令。

Answer 2

你想運行，直到他們鍵入正確的東西？這很簡單。你應該使用一個循環。

循環通常是一個重複自己，直到條件滿足爲止的語句塊。例如，對你來說一個好的循環是：

Repeat the program until the string equals Door or Candle。

條件，我看你已經很親近了。然後，讓我向您介紹while循環的塊的語法，我認爲它最適合您的情況（do while是適合您的另一個循環，但解釋稍微複雜一點）。 編輯：我改變雨水的評論代碼：

string startout=""; // Initiallizing it before the while loop, so it could be used inside it. 
cout << "You wake up in a room. There is a small lit candle and a door. What do you do?" << endl; 
while(startout !="door" && startout !="candle") // Do the following actions in this block until startout equals "Door" or "Candle". 
{ // start of a while block 
    getline(cin, startout); 
    if (startout.tolower().find("door")!=npos) // If the string contains any variation of door: 
    { 
     cout << " " << endl; 
     cout << "You move to the door but its too dark to see anything besides the outline of it." << endl; 
    } 
    else if (startout.tolower().find("candle")!=npos) // same, with candle. Much more elegant, like thomas said. 
    { 
      cout << "You pick up the candle then move to the door. With the light from the candle you can see the door well. What do you do?" << endl; 
    } 
    else 
    { 
     cout << "I don't understand " << startout << ". What do you do?" << endl; // If you didn't get what you want, ask for another string and restart. 
    } 
} // end of the while block. 

Answer 3

for (;;) 
    cout << "You wake up in a room. There is a small lit candle and a door. What do you do?" << endl; 
    string startout; 
    getline(cin, startout); 
    if (startout == "Door" || startout == "DOOR" || 
     startout == "door" || startout == "Open Door" || 
     startout == "open door" || startout == "OPEN DOOR" || 
     startout == "Open door" || startout == "Open the door" || 
     startout == "open the door") { 

     cout << " " << endl; 
     cout << "You move to the door but its too dark to see anything besides the outline of it." << endl; 
     break; 
    } else if (startout == "Candle" || startout == "CANDLE" || 
       startout == "candle" || startout == "Pick Up Candle" || 
       startout == "PICK UP CANDLE" || 
       startout == "pick up candle" || 
       startout == "Pick up candle") { 
     cout << "You pick up the candle then move to the door. With the light from the candle you can see the door well. What do you do?" << endl; 
     break; 
    }; 
} 

不是最優雅的方式來做到這一點，但你的想法。

Answer 4

首先，你需要一個更好的解析器。至少，將 合法值放在表格或某種地圖中，並指向 操作。然後換行輸入的功能：

typedef void (*  ActionPointer)(); // Or whatever you need. 
typedef std::map< std::string, ActionPointer, CaseInsensitiveCmp > 
        ActionMap; 

ActionPointer 
getAction(ActionMap const& legalValues) 
{ 
    std::string line; 
    if (! std::getline(std::cin)) { 
     // Error on std::cin... Probably fatal. 
    } 
    ActionMap::const_iterator action = legalValues.find(); 
    return action == legalValues.end() 
     ? nullptr 
     : action->second; 
} 

然後，你可以寫類似：

std::cout << "You wake up in a room. There is a small lit candle and a door." 
      " What do you do?" << std::endl; 
ActionPointer nextAction = getAction(); 
while (nextAction == nullptr) { 
    std::cout << "I don't understand. What do you do?" << std::endl; 
    nextAction = getAction(); 
} 
(*nextAction)(); 

如果你想在錯誤信息的更多信息，你可以安排 回報struct與指針和那個信息（還在測試 的指針）。