满足中断条件后循环继续

Question

我当前的任务要求我允许用户传递字符串并将罗马数字转换为十进制值。我遇到了一个问题，当随机单词通过程序时，它必须返回最长的有效前缀并忽略剩余的字符。我的代码通过一个while循环运行，一旦达到中断条件，它就会再次为剩余的字符重复代码。我不确定下一步应该采取什么措施来修复此代码以正确输出。下图显示了要通过代码的字符。 Image of test failure

#include <iostream>
#include <ctype.h>

using namespace std;
int
main (){
   while (!cin.eof ()){
    char inputChar;
    int result = 0;
    while (cin.get (inputChar))
    {
        inputChar = toupper (inputChar);
        if (inputChar == 'M')
            result = result + 1000;
        else if (inputChar == 'D')
        {
            inputChar = cin.peek ();
            inputChar = toupper (inputChar);
            if (inputChar == 'M')
            {
                result = result - 500;
                continue;
            }
            else if (inputChar == 'D'){
                result = result;
                continue;
            }
            else
            {
                result = result + 500;
                continue;
            }
        }
        else if (inputChar == 'C')
        {
            inputChar = cin.peek ();
            inputChar = toupper (inputChar);
            if (inputChar == 'M' || inputChar == 'D')
            {
                result = result - 100;
                continue;
            }
            else
            {
                result = result + 100;
                continue;
            }
        }
        else if (inputChar == 'L')
        {
            inputChar = cin.peek ();
            inputChar = toupper (inputChar);
            if (inputChar == 'M' || inputChar == 'D' || inputChar == 'C')
            {
                result = result - 50;
                continue;
            }
            else
            {
                result = result + 50;
                continue;
            }
        }
        else if (inputChar == 'X')
        {
            inputChar = cin.peek ();
            inputChar = toupper (inputChar);
            if (inputChar == 'M' || inputChar == 'D' || inputChar == 'C'
                || inputChar == 'L')
            {
                result = result - 10;
                continue;
            }
            else
            {
                result = result + 10;
                continue;
            }
        }
        else if (inputChar == 'V')
        {
            inputChar = cin.peek ();
            inputChar = toupper (inputChar);
            if (inputChar == 'M' || inputChar == 'D' || inputChar == 'C'
                || inputChar == 'L' || inputChar == 'X')
            {
                result = result - 5;
                continue;
            }
            else
            {
                result = result + 5;
                continue;
            }
        }
        else if (inputChar == 'I')
        {
            inputChar = cin.peek ();
            inputChar = toupper (inputChar);
            if (inputChar == 'M' || inputChar == 'D' || inputChar == 'C'
                || inputChar == 'L' || inputChar == 'X' || inputChar == 'V')
            {
                result = result - 1;
                continue;
            }
            else
            {
                result = result + 1;
                continue;
            }
        }
        else{
            break;
        }
    }
    if (result != 0)
    {
        cout << result << endl;
    }
}
return 0;
}

Answer 1

也许尝试在外循环结束时清除缓冲区。这样，最长的前缀就会被打印出来，并且不会在迭代中再次使用相同的单词。

    if (result != 0)
    {
        cout << result << endl;
    }
    cin. ignore(10000, '\n'); //used to ignore or clear characters from the input buffer
}

Answer 2

您的break 语句不会退出两个while 循环。它只打破了内部循环，而不是外部循环。您可以使用标志告诉外部循环不要继续：

int main() 
{
   bool exit_loop = false; // define a flag here
   while (!cin.eof()) {
      char inputChar;
      int result = 0;
      if (exit_loop) // check the flag here
         break;
      while (cin.get(inputChar))
      {

在打破时将平面设置为真：

 }
 else 
 {
    exit_loop = true; // set the flag here
    break;
 }