C++菜單中的循環

Question

在我的代碼中，即使輸入'Q'或'q'，程序仍然會循環顯示菜單。這裏有什麼問題？下面是代碼：

{ 
    char selection; 
    do { 
     cout << "Add a county election file   A" << endl; 
     cout << "Show election totals on screen  P" << endl; 
     cout << "Search for county results   S" << endl; 
     cout << "Exit the program     Q" << endl; 
     cout << "Please enter your choice: "; 
     cin >> selection; 
    } while ((selection != 'Q' || selection != 'q')); 
    return 0; 
} 

Answer 1

你想使用AND（&&）運營商在測試，而不是或（||）運算符。否則，selection != 'Q'和selection != 'q'之一將始終爲真，並且您的循環將永遠不會退出。

Answer 2

正如指出的那樣，||不符合您的要求。您需要使用&&運營商。

如果你按q，那就是這種情況。

(selection != 'Q' || selection != 'q') 
|---------------| |--------------| 
    true     false 

如果你按Q，那就是這種情況。

(selection != 'Q' || selection != 'q') 
|---------------| |--------------| 
    false     true 

循環應該是這樣的。

while((selection != 'Q' && selection != 'q')); 

Answer 3

試試這個：

} while((selection != 'Q' && selection != 'q')); 

Answer 4

使用toupper()。

如果你這樣做，while (toupper(selection) != 'Q' )你不需要檢查大寫和小寫。

#include <iostream> 
#include <stdio.h> 
using namespace std; 

int main(void) 
{ 



    char selection; 
    do { 
     cout << "The Menu" << endl; 
     cout << "____________________________________" << endl; 
     cout << "Add a county election file   A" << endl; 
     cout << "Show election totals on screen  P" << endl; 
     cout << "Search for county results   S" << endl; 
     cout << "Exit the program     Q" << endl; 
     cout << "____________________________________" << endl; 
     cout << "Please enter your choice: "; 
     cin >> selection; 
    } while (toupper(selection) != 'Q' ); 



    cout<<" \nPress any key to continue\n"; 
    cin.ignore(); 
    cin.get(); 

    return 0; 
}