在 C++ 菜单中使用 While 循环切换大小写

Question

我一直在用 C++ 构建一个菜单驱动的控制台，我目前使用 switch-case 作为我的选项，但现在我陷入了 switch case。

场景如下:

SCENARIO

说明: 在主菜单中输入无效选项后，会出现错误，提示用户重新输入他们想要的选项，现在我的问题是用户第二次尝试输入正确的选项, 它循环回到主菜单而不是将其重定向到下一个菜单。

我的目标：不重新显示主菜单，直接从默认进入二级菜单。

我的部分代码:

#include <iostream>
#include <string>
#include <stdlib.h>

using namespace std;
int choice;
int booknumber;
int booktitle;
int author;
int datepublished;
int e = 0;

void menu();
void inputbook();
void searchbook();
void borrowbook();
void exit();

//CLASS
class Books
{
    public:
        int booknumber;
        string booktitle;
        string author;
        string datepublished;
        Books(const int booknumber, const string booktitle, const string author, const string datepublished) : booknumber(booknumber), booktitle(booktitle), author(author), datepublished(datepublished) {}
};

//MAIN
int main()
{ 
    while (true)
    {
        cout << endl;
        if (e == 1)
        {
            break;
        }
        menu ();
    }

    return 0;    
}

//MENU
void menu()
{
    cout << "Welcome to DLC Library System\n";
    cout << "Final Project in Advance Programming\n\n";

    cout << "PROGRAMMER\n";
    cout << "ME\n\n";

    cout << "====================================\n";
    cout << "[1] -------- Input Book ------------\n";
    cout << "[2] -------- Search Book -----------\n";
    cout << "[3] -------- Borrow Book -----------\n";
    cout << "[4] -------- Exit Program ----------\n";
    cout << "====================================\n";
    cout << "Input your choice (Number Only): ";
    cin >> choice;

    switch (choice)
    {
    case 1:
        inputbook ();
        break;
    case 2:
        searchbook ();
        break;
    case 3:
        borrowbook ();
        break;
    case 4:
        exit();
        break;
    default:
        while (choice < 1 || choice > 4)
        {
            cout << "Wrong Option\n";
            cout << "Input your choice (Number Only): ";
            cin >> choice;

                if (choice < 1 || choice > 4)
            {
                continue;
            }
        }
    }
    
    
}

// INPUT BOOK
void inputbook ()
{
    int booknumber;
    string booktitle;
    string author;
    string datepublished;

    cout << "INPUT NEW BOOK\n\n";

    cout << "Book Number: \n";
    cin >> booknumber;

    cout << "Book Title: \n";
    cin >> booktitle;

    cout << "Author: \n";
    cin >> author;

    cout << "Date Publish: \n";
    cin >> datepublished;
    Books(booknumber,booktitle, author, datepublished);
    
    cout << "====================================\n";
    cout << "[1] -------- Try Again? ------------\n";
    cout << "[2] -------- Return to Menu --------\n";
    cout << "[3] -------- Exit Program ----------\n";
    cout << "====================================\n";
    cout << "Input your choice (Number Only): ";
    cin >> choice;

    switch (choice)
    {
    case 1:
        inputbook ();
        break;
    case 2:
        menu ();
        break;
    case 3:
        exit();
    default:
        cout << "Wrong Option";
    }
}

Answer 1

最好避免重复代码。在这里，您有一个默认情况，它本质上是一个输入循环，而您可以在开始时完成该输入循环。所以按照你写的方式，你仍然需要一个围绕整个事情的循环，加上更多的逻辑，这使得代码更难阅读和更多 bug-prone.

为什么不简单：

cout << "====================================\n";
cout << "[1] -------- Input Book ------------\n";
cout << "[2] -------- Search Book -----------\n";
cout << "[3] -------- Borrow Book -----------\n";
cout << "[4] -------- Exit Program ----------\n";
cout << "====================================\n";

int choice;
bool validInput = false;

while (!validInput)
{
    cout << "Input your choice (Number Only): ";
    if (!(cin >> choice)) {
        std::cerr << "Aborted\n";
        return;
    }

    validInput = (choice >= 1 && choice <= 4);
    if (!validInput) {
        std::cout << "Invalid input\n";
    }
}

switch(choice)
{
    // ...
}

现在，如果您愿意，可以让您的输入例程更加健壮。请注意，如果输入失败，我已经退出了该函数。这可能来自流错误，但也可能是用户输入了 non-integer 值。

您可能希望使用 std::getline 将输入读取为字符串，然后使用 std::stoi 将其转换为整数或解析 std::istringstream.[=14 中的值=]

Answer 2

改为

continue;

试着打电话

inputbook();

所以它不会回去。这是您面临的问题，因为您再次调用了 switch-case，即“继续”。这就是为什么当用户输入您刚刚设置的可接受的 int 范围时它返回菜单的原因。

Answer 3

只需修改如下代码，在进入开关前处理有效输入验证，这样就可以简单的缓解你遇到的问题！

cin >> choice;
while (choice < 1 || choice > 4)
    {
        cout << "Wrong Option\n";
        cout << "Input your choice (Number Only): ";
        cin >> choice;

            if (choice < 1 || choice > 4)
        {
            continue;
        }
    }

switch (choice)
{
case 1:
    inputbook ();
    break;
case 2:
    searchbook ();
    break;
case 3:
    borrowbook ();
    break;
default:
    exit();
    break;
}

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