Question

每次我运行代码并故意输入错误的输入以获取输出Please enter a valid choice!时，它都会执行，但随后程序结束。

这是一个模拟自动售货机的C ++程序。我正在使用Visual Studios。我尝试在开关外使用while循环，使用for循环，使用if语句，以及其他我不太确定的东西（使用我在google上找到的东西）。

#include "pch.h"
#include <iostream>
#include <string>
#include <cctype>
#include <iomanip>
using namespace std;

struct Drinks
{
    string name;
double cost;
int number;
};


class Simulation
{
private:
    Drinks softDrink[5];
    double money;
bool enoughMoney;
int amount;
double total;
void dailyReport();
void inputMoney(int);

public:
Simulation();
void displayChoices();
void buyDrink(int);
~Simulation();


};

Simulation::Simulation()
{
for (int x = 0; x < 5; x++)
{
    if (x == 4)
    {
        softDrink[x].cost = 1.50;
    }
    else
    {
        softDrink[x].cost = 1.00;
    }

    softDrink[x].number = 20;
}

softDrink[0].name = "Cola";
softDrink[1].name = "Root-beer";
softDrink[2].name = "Orange soda";
softDrink[3].name = "Grape soda";
softDrink[4].name = "Bottled water";

money = 0.0;
amount = 0;
total = 0.0;
}
void Simulation::displayChoices()
{
cout << fixed << setprecision(2);
for (int x = 0; x < 5; x++)
{
    cout << x + 1 << "." << softDrink[x].name << " : $" << softDrink[x].cost << endl;
}
}
void Simulation::inputMoney(int choice)
{
cout << "The cost for " << softDrink[choice - 1].name << " is $" << softDrink[choice - 1].cost << endl;
cout << "How much money will you insert? ";
cin >> money;

if (money < softDrink[choice - 1].cost)
{
    cout << "Can not deliver beverage, not enough money" << endl;
    enoughMoney = false;
}
else
{
    enoughMoney = true;
}
}
void Simulation::buyDrink(int choice)
{
char purchaseChoice;

inputMoney(choice);

if (enoughMoney)
{

    cout << "Do you want to make a purchase for this drink?(Y/N) ";
    cin >> purchaseChoice;
    while (purchaseChoice != 'Y' && purchaseChoice != 'N')
    {
        cout << "Please enter correct input, purchase drink (Y/N)? ";
        cin >> purchaseChoice;
    }


    if (purchaseChoice == 'Y')
    {
        if (softDrink[choice - 1].number <= 0)
        {
            cout << "We are sold out" << endl;
            cout << "Here is your money back $" << money << endl;
        }
        else
        {
            cout << "Here is your beverage" << endl;
            total += softDrink[choice - 1].cost;
            softDrink[choice - 1].number -= 1;
        }


        if (money > softDrink[choice - 1].cost)
        {
            money -= softDrink[choice - 1].cost;
            cout << "Here is your change $" << money << endl;
        }
    }
    else
    {
        cout << "Here is your money back $" << money << endl;
    }
}
}
Simulation::~Simulation()
{
for (int x = 0; x < 5; x++)
{
    cout << "There are " << softDrink[x].number << " " << softDrink[x].name << " left in the machine" << endl;
}

cout << "The total amount of money collected during the day is $" << total << endl;
}



int main()
{
int choice;
string line;
Simulation machine;

cout << fixed << setprecision(2);


do
{
    cout << endl << "Drink Machine Menu" << endl;
    machine.displayChoices();
    cin >> choice;

    switch (choice)
    {
        case 1: machine.buyDrink(choice);
            break;

        case 2: machine.buyDrink(choice);
            break;

        case 3: machine.buyDrink(choice);
            break;

        case 4: machine.buyDrink(choice);
            break;

        case 5: machine.buyDrink(choice);
            break;
        default: cout << endl << "Please enter a valid choice!" << endl << endl;
            cin >> choice;
    }
} while (choice != 6 && choice >= 1 && choice <= 5);
return 0;
}

我希望它不断告诉用户Please enter a valid choice!，但是它只输出一次然后退出程序

Answer 1

您可以将while条件更改为

while (choice != 6);

，除非输入6，否则它不会退出循环（如果要在输入时退出）。其余部分在交换机中处理，您无需再检查一下。

如何解决我的代码避免过早退出？

1 个答案: