更新＃1

Question

我知道我的方法可能有点尴尬，我正在尝试验证菜单的输入，以便只输入1-6之间的数字，并且不接受任何其他内容。我有工作代码，我把输入作为一个字符串，然后将其更改为一个int，以便在切换案例中使用它，但我知道我可以使它更有效地工作。有什么想法吗？

 void menu(double pi, char ssTwo)                                                           //menu for choosing a shape
{
    string choice;

    cout << "Welcome to the shape calculator!\n\nPlease select what you wish to calculate:\n\n1 - Area of a Circle\n\n2 - Circumference of a Circle\n\n3 - Area of a Rectangle\n\n4 - Area of a Triangle\n\n5 - Volume of a Cuboid\n\n6 - Exit the program\n\n ";
    cin >> choice;

    while (choice != "1" && choice != "2" && choice != "3" && choice != "4" && choice != "5" && choice != "6")
    {
        cout << "Invalid input, please enter a number of 1-6\n\n";
        cin >> choice;

        cin.clear();
        cin.ignore(numeric_limits<streamsize>::max(), '\n');
    }

    int choiceInt = atoi(choice.c_str());
    system("CLS"); 

    switch (choiceInt)                                                          //switch case for each shape
    {
    case 1:
        circleArea(pi, ssTwo);
        break;
    case 2:
        circleCircum(pi, ssTwo);
        break;
    case 3:
        rectanArea(ssTwo);
        break;
    case 4:
        triangArea(ssTwo);
        break;
    case 5:
        cubVol();
        break;
    case 6:
        exitSystem();
        break;

    default:
        cout << "Invalid input, please enter a number of 1-5\n\n";
        menu(pi, ssTwo);
        break;
    }
}

Answer 1

将char视为Ascii值;查看http://rextester.com/XKMJ90988

#define ONE 49
#define TWO 50
int main()
{
    char val = '1';

    switch (val)
    {
        case ONE:
        {
            std::cout << "1 was Selected";
            break;
        }
        case TWO :
        {
            std::cout << "2 was Selected";
            break;
        }   

        default:
            std::cout << "Invalid input, please enter a number of 1-5\n\n";
            break;
    }
}

更新＃1

检查：http://rextester.com/WWUM1166

如果你将cin中的值读入char而不是字符串，它只会占用第一个字符;因此，如果用户按'1asdfad'，您将只在char中使用1，您将使用1;如果按下'asdfasdf'，它将显示'a'并显示无效输入;

int main()
{
    char val;

    std::cout << "Enter a number: ";
    std::cin >> val;

    std::cout << val << "\n";

    switch (val)
    {
        case '1':
        {
            std::cout << "1 was Selected";
            break;
        }
        case '2':
        {
            std::cout << "2 was Selected";
            break;
        }   

        default:
            std::cout << "Invalid input, please enter a number of 1-5\n\n";
            break;
    }
}

操纵字符串成1个字符

1 个答案:

更新＃1