C++如何重新启动if语句？

Question

说你的代码是：

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; 
      }; 

如果使用的代码类型的东西，除了“门”或“蜡烛”（或任何变体），还可以将代码是什么重启if语句，因此重新谁是 - 问第一个问题？

例如：用户输入：“跳舞”

输出：“我不明白，‘跳舞’。 你是做什么？”

或类似的东西。

Answer 1

您可以在外部添加一个while(true)循环，并添加if条件，该条件在正确输入时会中断。

另外，函数getline命令之后，如果别的conditon之后添加其他条件：

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

在其他if和else if子句，添加一个break;命令。

Answer 2

你想运行，直到他们键入正确的东西？这很简单。你应该使用一个循环。

循环通常是一个重复自己，直到条件满足为止的语句块。例如，对你来说一个好的循环是：

Repeat the program until the string equals Door or Candle。

条件，我看你已经很亲近了。然后，让我向您介绍while循环的块的语法，我认为它最适合您的情况（do while是适合您的另一个循环，但解释稍微复杂一点）。 编辑：我改变雨水的评论代码：

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; 
    }; 
} 

不是最优雅的方式来做到这一点，但你的想法。

Answer 4

首先，你需要一个更好的解析器。至少，将 合法值放在表格或某种地图中，并指向 操作。然后换行输入的功能：

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; 
} 

然后，你可以写类似：

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)(); 

如果你想在错误信息的更多信息，你可以安排 回报struct与指针和那个信息（还在测试 的指针）。