在使用数据结构从文件中读取数据时遇到麻烦

我有一个包含汽车租赁信息和公司信息的程序。我正试图读取它，并有它显示在终端上。然而，我只让一家公司打印清晰，而其他公司只打印垃圾。我也想用5辆车的库存来存储代理，但不知道如何在没有读取所有信息的情况下存储它们。我也只能使用c风格的字符串。

这是我正在阅读的文件:

Hertz 93619
2014 Toyota Tacoma 115.12 1
2012 Honda CRV 85.10 0
2015 Ford Fusion 90.89 0
2013 GMC Yukon 110.43 0
2009 Dodge Neon 45.25 1
Alamo 89502
2011 Toyota Rav4 65.02 1
2012 Mazda CX5 86.75 1
2016 Subaru Outback 71.27 0
2015 Ford F150 112.83 1
2010 Toyota Corolla 50.36 1
Budget 93035
2008 Ford Fiesta 42.48 0
2009 Dodge Charger 55.36 1
2012 Chevy Volt 89.03 0
2007 Subaru Legacy 59.19 0
2010 Nissan Maxima 51.68 1

需要帮助的代码段:

#include <iostream>
#include <fstream>
using namespace std;
struct car
{
        char agency[10];
        int zip;
        int year;
        char make[10];
        char model[10];
        float price;
        int available;
} ;
struct agency
{
    char company[10];
    int zip;
    int inventory[5];
};
void menu();

// Main Function
int main ()
{
    // declare variables
    const int carAmount = 15;
    int agencyAmount = 1;
    int choice;
    agency agencyLib[carAmount];
    car carLib[carAmount];
    char filename[10];
    ifstream carInData;
    bool menu1 = false;
    //prompt user for input file
    cout << " Enter file name: ";
    cin >> filename;
    // Start loop menu
    while(menu1 = true)
    {
        menu();
        carInData.open(filename);
        cin >> choice;
        if (carInData.is_open())
        {
            // read list of names into array
            for (int count = 0; count < agencyAmount; count++) 
            {
                carInData >> agencyLib[count].company >> agencyLib[count].zip;
                for (count = 0; count < carAmount; count++)
                {
                    carInData >> carLib[count].year >> carLib[count].make >> carLib[count].model >> carLib[count].price >> carLib[count].available;
                }
            }
        }
    switch (choice)
    {
        // Case 1 closes menu
        case 1:
            return 0;
            break;
        // Case 2 displays if car is available if 1, unavailable if 0
        case 2:
        // itterate through car array
            for(int count = 0; count < agencyAmount; count++)
            {
                cout << agencyLib[count].company << " " << agencyLib[count].zip << "n";
                for(int count = 0; count < carAmount; count++)
                {
                    // Displays if car is available or not 
                    /*      if (carLib[count].available == 1)
                    cout << " Available ";
                    else
                        cout << " Unavailable ";
                    */
                    // Display Cars
                    cout << carLib[count].year << " " << carLib[count].make << " " << carLib[count].model << " " << carLib[count].price << "  " << "n";
                }
            }
        }
    }
}