Question

#include "Menu.h"
#include <iostream>
#include <stdio.h>
#include <string>


void Menu::displayMenu() {
    cout << " 1. New User login \n 2. View available vehicles \n 3. Return Vehicle \n 4. Exit Program \n";

    char choice;
    cin >> choice;

main:
    switch (choice) {
    case '1': //new user option
        login();
        break;


    case '2': //This will be the option for viewing available cars. each choice in this switch will lead to car info and choice to rent.

    carmenu:
        cout << "Now viewing available vehicles...Select vehicle for further information...\n";
        cout << " 1. Toyota Carolla \n 2. Toyota Rav4 \n 3. Toyota Prius \n 4. Ford Mustang \n 5. Ford Transit \n 6. Main Menu \n";
        char rent;
        char carchoice;
        cin >> carchoice;
        switch (carchoice) {
        case '1':
            cout << "Toyota Carolla..\n Passenger class \n 30 MPG \n 4 Doors \n";
            cout << "Enter Y to rent vehicle or enter N to return to car selection...\n";
            cin >> rent;
            if (rent == 'Y') {
                //construct

            }
            else if (rent == 'N') { goto carmenu; };

        case '2':
            cout << "Toyota Rav4..\n Passenger class \n 24 MPG \n 5 Doors \n";
            cout << "Enter Y to rent vehicle or enter N to return to car selection...\n";
            cin >> rent;
            if (rent == 'Y') {
                //call passenger car constructor called Rav4 with proper data entries.

            }
            else if (rent == 'N') { goto carmenu; };
            break;

        case '3':
            cout << "Toyota Prius.. \n Passenger class \n 54 MPG \n 4 Doors \n";
            cout << "Enter Y to rent vehicle or enter N to return to car selection...\n";
            cin >> rent;
            if (rent == 'Y') {
                //call passenger car constructor called Prius with proper data entries.

            }
            else if (rent == 'N') { goto carmenu; };
            break;

        case '4':
            cout << "Ford Mustang.. \n Passenger class \n 18 MPG \n 2 Doors \n";
            cout << "Enter Y to rent vehicle or enter N to return to car selection...\n";
            cin >> rent;
            if (rent == 'Y') {
                //call passenger car constructor called Mustang with proper data entries.
                break;
            }
            else if (rent == 'N') { goto carmenu; };
            break;

        case '5':
            cout << "Ford Transit ..\n Cargo class \n 14 MPG \n";
            cout << "Enter Y to rent vehicle or enter N to return to car selection...\n";
            cin >> rent;
            if (rent == 'Y') {
                Cargo van(14);
                van.set_purpose();

            }

            else if (rent == 'N') { goto carmenu; };

        case '6':
            goto main;


        }//second switch

    }//first switch
}//function end

我希望这不太难以阅读......我正在做一个项目，我必须为公司设计一个汽车租赁计划。主菜单的第二种情况将用户带到一个菜单，在该菜单中他可以选择要租用的汽车类型。菜单还没有完成，但我已经完成了汽车选择菜单选项5的课程。租用汽车时（为if语句输入Y）正确调用货物的构造函数，但程序要么在输入目的后立即结束，要么返回到汽车租赁页面。我不知道在哪里放假或去发言。任何人都可以帮忙??

Answer 1

通过将break放在每个案例的末尾，它将突破当前案例下面的案例并执行其余代码。如果您为租金选择输入“Y”，您的程序将再次返回到汽车菜单，因为变量choice永远不会被重新分配，因此它将保持为“2”。如果输入“N”，您将返回选择汽车，因为goto carmenu。对于所有其他输入，您没有其他情况，因此程序将在中断后退出。

我建议将代码分解为函数，而不是使用goto。如果您遇到这样的错误，可以更容易理解和理解。

在嵌套开关内断开和转到放置

1 个答案: