Question

我是一年级CS专业。今天，在我们的实验室中，我们不得不调试一些代码并使之工作。结果如下。

#include <iostream>

using namespace std;

int main() {

   int x = 3, y;
   char myanswer;
   int val= 1;
   int num;

   y = x;
   cout << "y is set to: " << y << endl;


   bool again = true;
   int ans;
   while (again) {
      cout << "Please input a number: ";
      cin >> y;
      if (x > y)
         cout << "X is greater than Y\n";

      else {
         cout << "X is less than Y" << endl;
         cout << "would you like to input another number?" << endl;
         cin >> ans;
         if (ans != 1)
            break;
      }
      cout << "would you like to input another number ?" << endl;
      cin >> ans;
      if (ans != 1)
         again = false;
   }
   for (x = 0; x < 10; x++)
      cout << x << endl;

   cout << "What number would you like to find the factorial for? " << endl;


   cin >> num;
   cout << num;
   for (int x = num; x > 0; x--) {
      val *= x;
   }
   cout << "Are you enjoying cs161? (y or n) " << endl;

   cin >> myanswer;
   if (myanswer == 'y')
      cout << "Yay!" << endl;
   else
      cout << "I hope you will soon!" << endl;

   return 0;

}

在关于阶乘的判例后，cin无效，并且用户不再能够输入输入。到目前为止，我的实验室和朋友还没有发现问题。该代码已在我学校的工程服务器和本地计算机上进行编译和提取。在这两个错误上仍然存在。

Answer 1

几乎可以肯定这会导致溢出

   for (int x = num; x > 0; x--) {
      val *= x;
   }

您输入的数字是什么？

Answer 2

当您声明为：

cout << "would you like to input another number?" << endl;

用户的第一个直觉是输入y或n作为答案。您可以通过提供提示来帮助用户。

cout << "would you like to input another number (1 for yes, 0 for no)?" << endl;

如果这样做，最好在整个程序中保持一致。寻求y / n响应的下一个提示必须使用相同的机制。

cout << "Are you enjoying cs161? (1 for yes, 0 for no) " << endl;

当然，在继续使用数据之前，请务必先验证输入操作。

if ( !(cin >> ans) )
{
   // Input failed. Add code to deal with the error.
}

为什么我的两个cin语句没有在程序结束时运行？

2 个答案: