Question

您好我正在处理工资申请，其中有四个选项，如下所示：

从商店银行帐户中减去一笔金额（该帐户是文本文件“shop”）。您无需添加相同的金额另一个帐户，但您应该在单独的文件中记录带有时间戳的事务。该申请应防止帐户余额被透支。
列出屏幕上最近的五笔交易。如果还没有五笔交易然后列出所有这些。
将帐户名称，号码和当前余额打印到屏幕上。
退出程序。

我已经完成了1,3和4但我完全不知道如何绕过2号。我希望有人能指出我正确的方向。

#include <limits>
#include "stdafx.h"
#include <iostream>
#include <fstream>
#include <ctime>
#include <string>


int  read_balance(void);
void write_balance(int balance);

using namespace std;

int _tmain(int argc, _TCHAR* argv[])
{
    int selection;
    int total;
    int attempts = 0;
    string name;
    string number;



    cout << "Correct login details entered!" << "" << endl;
    cout << "1. Transfer an amount" <<endl;
    cout << "2. List recent transactions"<<endl;
    cout << "3. Display account details and current balance"<<endl;
    cout << "4.Quit" << endl;
    cout << "Please enter menu number"<<endl;
    cin >> selection;


    switch(selection)
    {
    case 1: 
        cout << "How much do you wish to transfer?" << endl;
        int amount = 0;

        if (std::cin >> amount)
        {
            std::cout << "Transferred Amount:" << amount << "\n";
            int balance = read_balance();

            if (amount <= 0)
            {
                std::cout << "Amount must be positive\n";
            }
            else if (balance < amount)
            {
                std::cout << "Insufficient funds\n";
            }
            else
            {
                int new_balance = balance - amount;

                write_balance(new_balance);
                std::cout << "New account balance: " << new_balance << std::endl;

                fstream infile("time.txt", ios::app);



                std::time_t result = std::time(nullptr);
                std::string timeresult = std::ctime(&result);


                infile << amount << std::endl;
                infile << timeresult << std::endl;
            }

        }
    break;


    case 2:
        cout << "Here are you're recent transactions" <<endl;
    break;

    case 3:
        cout << "The account names is:" << name << endl;
        cout << "The account number is:" << number << endl;
        std::cout << "The current account balance is " << read_balance() << std::endl;
    break;

    case 4:
        system("pause");
        return 0;

    default:
        cout << "Ooops, invalid selection!" << endl;
        break;

    }

    system("pause");
    return 0;
}


int read_balance(void)
{
    std::ifstream f;
    f.exceptions(std::ios::failbit | std::ios::badbit);
    f.open("shop.txt");
    int balance;
    f >> balance;
    f.close();
    return balance;
}

void write_balance(int balance)
{
    std::ofstream f;
    f.exceptions(std::ios::failbit | std::ios::badbit);
    f.open("shop.txt");
    f << balance;
    f.close();
}

Answer 1

你可以说一下交易清单：

std::list<std::pair<int,std::string> > last_transactions

编写一个函数，每次处理新事务（pair）时都会在列表的末尾插入新的push_back，如果列表中已有6个项目，它将删除最旧的一个（前面的那个 - pop_front）。您可以使用简单的for循环和迭代器来列出此事务。

将事务从文本文件输出到屏幕

1 个答案: