switch 语句 C++ [默认部分]

Question

我正在编写一个关于 switch 语句的基本程序。 5个选项的菜单，如果用户输入无效数字，应再次提示用户进行选择（范围为1-5）。到目前为止我得到了什么：

#include <iostream>
#include "Menu.h"
using namespace std;

void Menu::inputChoices()
{
    int choice;

    cout << "IHCC Computer Science Registration Menu" << endl;
    cout << "1. Welome to Computer Programming in C++" << endl;
    cout << "2. Welcome to Java Programming" << endl;
    cout << "3. Welcome to Android Programming" << endl;
    cout << "4. Welcome to iOS Programming" << endl;
    cout << "5. Exit" << endl;
    cout << "\nEnter your selection: " << endl;

    while ((choice = cin.get()) != EOF)
    {
        switch (choice)
        {
        case '1':
            cout << "Welcome to Computer Programming in C++" << endl;
            break;
        case '2':
            cout << "Welcome to Java Programming" << endl;
            break;
        case '3':
            cout << "Welcome to Android Programming" << endl;
            break;
        case '4':
            cout << "Welcome to iOS Programming" << endl;
            break;
        case '5':
            cout << "Exiting program" << endl;
            break;
        default:
            cout << "Invalid input. Re-Enter your selection: " << endl;
        }
    }
}

这个项目有3个文件，这是源文件。我的问题是当我输入的数字在 (1-5) 范围内时，switch 的默认部分仍然会显示出来。我只想显示我的选择。谁能帮帮我！非常感谢

Answer 1

应该处理回车键'\n'。请尝试以下操作，看看是否有效：

while ((choice = cin.get()) != EOF)
{
  case 1:
  // ...
  // Basic idea is read extra `\n` and ignore it. Use below if condition
  default:
        if(choice != '\n')
          cout << "Invalid input. Re-Enter your selection: " << endl;
}

Answer 2

您还获得默认值的原因是因为cin.get() 正在读取换行符 (10)。

我认为在这里使用cin.get 没有意义，因为输入23987928 或asdasf 之类的内容会输出大量行...您应该使用cin >> choice 并将您的案例转换为ints .比如：

cin >> choice;
switch (choice)
{
    case 1:
        cout << "Welcome to Computer Programming in C++" << endl;
        break;
    // ...
    default:
        cout << "Invalid input. Re-Enter your selection: " << endl;
}