Cin 没有得到正确的输入

Question

我用 C++ 编写的程序有问题。请注意，我不在笔记本电脑上（接下来的 8 小时我都在工作），所以我无法在此处复制代码，但我在这个问题上花了几个小时，所以我知道出了什么问题，只是不要不知道如何解决它。所以我的代码看起来像这样：

bool valid = false;
int choice;

cout << "Please make a choice between 1 and 4\n"; 
//Here I enumarate 4 choices
cin >> choice;  //first input

//note that I do verify that an integer is entered and that it's between 1 and 4
//my code works so far

valid = true; //if the choise is good

while(valid)
{
     if (choice == 1)
     {
         //here I open a url
         cout << "If you would like to stick with this choice, presse 1 again or make  another choice \n";

         //This is where the code does not work at all
         cin >> choice; //second input

         if (choice == 1) //this line is to get out of the loop but I never reach it
             valid = false;
     }
     if (choice == 2) //contains similar code to the 1st if
     {
     }
     if (choice == 3) //contains similar code to the 1st if
     {
     }
     if (choice == 4) //contains similar code to the 1st if
     {
     }
     else
     {
          cout << "This is not one the choices. Choose again!";
          cin >> choice;
     }
}

当用户选择第一个选项时，我只填写了第一个 if，因为其他 3 个 if 的行为方式相同，只是使用了不同的链接。以下是用户选择第一个选项时程序的行为方式。它可以很好地打开链接，然后要求用户再次按 1 以坚持使用此选项或做出其他选择。它总是在后面进入 else 条件。即使我按 1 退出循环，它也会进入 else。这意味着无论我输入什么输入，它都没有注册为 1 到 4 之间的整数。

我已尝试调试以查看“选择”在第二次输入之后的值，它显示为 49！这没有任何意义，因为我输入了 1。但是，在 else 之后，如果我在 1 和 4 之间做出正确的选择，例如 1，链接将再次打开就好了，程序将再次要求我按 1 坚持与我的选择或做出另一个选择。

我认为缓冲区上还有一些东西。我读了一些关于它以了解如何清除它，但我在第一个 cin 之后尝试了 cin.clear()、cin.ignore()、cin.sync() 但第二个输入总是与我实际输入的不同无论我输入什么，它都会再次进入 else 。这是一个来自地狱的无限循环。

我还是 C++ 新手，我正在尝试一些想法。我尝试自己解决这个问题，但我需要帮助。谢谢。

Answer 1

//Your control block was not correct. 
//Your logic falls into the else block after the second input 
//has been read because it use the new value of choice and run through all the
//if statements. In a case where you have multiple options in that manner use 'else if'` 

void start()
{
    bool valid = false;
    int choice;

    cout << "Please make a choice between 1 and 4\n"; 
    //Here I enumarate 4 choices
    cin >> choice;  //first input

    //note that I do verify that an integer is entered and that it's between 1 and 4
    //my code works so far

    valid = true; //if the choise is good

    while(valid)
    {
         if (choice == 1)
         {
             //here I open a url
             cout << "If you would like to stick with this choice, presse 1 again or make  another       choice \n";

             //This is where the code does not work at all
             cin >> choice; //second input

             if (choice == 1)
             {
                valid = false;
             } //this line is to get out of the loop but I never reach it

         }

         else if (choice == 2) //contains similar code to the 1st if
         {
         }
         else if (choice == 3) //contains similar code to the 1st if
         {
         }
         else if (choice == 4) //contains similar code to the 1st if
         {
         }
         else
         {
              cout << "This is not one the choices. Choose again!";
              cin >> choice;
         }
    }   
}
int main(void)
{
    start();
    return 0;
}

Answer 2

正如Codec所说，您的if else不正确，在您确认选择后，它会将您的选择与2、3和4的值进行比较。由于现在是1，它会进入else语句并执行那里的任何事情。我的建议是使用 switch 语句，这样会更容易阅读。

cin >> choice;

switch(choice)
{
    case 1:
    // do whatever you want
    break;
    case 2:
    // do whatever you want
    break;
    // continue with your choice
    default:
    // do whatever you want if the value falls out of your choice
    cout << "This is not the right choice" << endl;
    break;
}

Answer 3

在你的 while 循环中使用如下：

while(valid)
   {
    if(choice==1)
      {
        //your code
      }
else if(choice==2)
      {
       //your code
      }
else if(choice==3)
      {
       //your code
      }
else if(choice==4)
      {
       //your code
      }
else 
      {
       //your code
      }
}