switch語句C++ [默認部分]

Question

我在寫關於switch語句的基本程序。 5個選項菜單，如果使用輸入無效號碼，應再次提示用戶進行選擇（範圍從1-5）。在這裏我得到了什麼至今：

#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; 
     } 
    } 
} 

該項目有3個文件，這是源文件。我的問題是，當我輸入一個數字在（1-5）的範圍內，開關的默認部分仍然顯示。我只想顯示我的選擇。任何人都可以幫助我！非常感謝您

Answer 1

您之所以獲得默認值的原因是因爲cin.get()正在讀取換行符（10）。

我不使用cin.get認爲是有道理的，因爲在這裏進入像23987928或asdasf將輸出一噸的行...你應該使用cin >> choice和你的情況下，轉化成int秒。例如：

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

Answer 2

應該處理回車鍵'\ n'。嘗試下面，看看它的工作原理：

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; 
}