我在写关于switch语句的基本程序。 5个选项菜单,如果使用输入无效号码,应再次提示用户进行选择(范围从1-5)。在这里我得到了什么至今:switch语句C++ [默认部分]
#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;
}
}
}
该项目有3个文件,这是源文件。我的问题是,当我输入一个数字在(1-5)的范围内,开关的默认部分仍然显示。我只想显示我的选择。任何人都可以帮助我!非常感谢您
你好,如果其中一个答案帮助你解决了你的问题,请检查它是否正确。如果没有,我们可以尝试进一步帮助你。 – usandfriends 2015-05-29 00:59:22