How to deserialize and then put data into form?

Question

Goal

I currently have a valid method of saving the information

I then want the program to load the saved state upon clicking a 'load' button after exiting.

I then want the program to display the data in the appropriate place

In the form I have two DataGridView's one for an employee and another for a supervisor.

=============================================================

Method

I have serialised two generic lists into a .dat file

BinaryFormatter bFormatter = new BinaryFormatter();    
using (FileStream fs = new FileStream(FILENAME, FileMode.OpenOrCreate))
{
  bFormatter.Serialize(fs, employees);
  bFormatter.Serialize(fs, supervisors);
}

So far in the .dat file, I have 3 Employees and 2 Supervisors, and I am not sure how to extract the information and place them into the appropriate place

The lists are as follows:

List<Employee> employees = new List<Employee>();
List<Supervisor> supervisors = new List<Supervisor>();

Employee e1 = new Employee(MemberJob.Employee, "Name", MemberSkills.CPlus | MemberSkills.CSharp, false);
Supervisor s1 = new Supervisor(MemberJob.Supervisor, "Another name", false); 
employees.Add(e1);
supervisors.Add(s1);

=========================================================================

Attempt

I have had an extensive look around the internet and on Stackoverflow, but mainly it's irrelevant to my context or they are using the XML format which I do not want to use.

I assumed it would just be a case of replicating the serialize method and just changing the bFormatter.Serialize(fs, employees); to bFormatter.Deserialize(fs, employees); but I am stuck on what to do after I deserialize the list.

 BinaryFormatter bFormatter = new BinaryFormatter();
 FileStream fs = File.Open(FILENAME, FileMode.Open);
 object obj = bFormatter.Deserialize(fs);

The object then brings back the data I need but I cannot put the object obj data into a usable format, it is stuck in the object obj if possible I'd like to try to put it back into the Employee list

Answer 1

My Christmas present to you. Hope it helps. :)

namespace WpfApplication3
{
    public partial class App : Application
    {
        string path = @"C:\Users\xxx\Desktop\myFile.dat";

        public App()
        {
            InitializeComponent();

            //Test
            List<Employee> eList = new List<Employee>();
            eList.Add(new Employee("aaa"));
            eList.Add(new Employee("bbb"));

            List<Supervisor> sList = new List<Supervisor>();
            sList.Add(new Supervisor("ccc"));
            sList.Add(new Supervisor("ddd"));

            SavedInfo savedInfo = new SavedInfo();
            savedInfo.employeeList = eList;
            savedInfo.supervisorList = sList;


            SaveToFile(savedInfo); //Save to file

            SavedInfo newSaveGame = LoadFromFile(); //Load from file

            foreach (var e in newSaveGame.employeeList)
                Console.WriteLine("Employee: " + e.name);

            foreach (var e in newSaveGame.supervisorList)
                Console.WriteLine("Supervisor: " + e.name);
        }

        public void SaveToFile(SavedInfo objectToSerialize)
        {
            Stream stream = File.Open(path, FileMode.Create);
            BinaryFormatter bFormatter = new BinaryFormatter();
            bFormatter.Serialize(stream, objectToSerialize);
            stream.Close();
        }

        public SavedInfo LoadFromFile()
        {
            if (!System.IO.File.Exists(path))
                return new SavedInfo();

            SavedInfo objectToSerialize;
            Stream stream = File.Open(path, FileMode.Open);
            BinaryFormatter bFormatter = new BinaryFormatter();
            objectToSerialize = (SavedInfo)bFormatter.Deserialize(stream);
            stream.Close();
            return objectToSerialize;
        }
    }

    [Serializable()]
    public class SavedInfo
    {
        public List<Employee> employeeList = new List<Employee>();
        public List<Supervisor> supervisorList = new List<Supervisor>();
    }

    [Serializable()]
    public class Employee
    {
        public string name = "";

        public Employee(string eName)
        {
            name = eName;
        }
    }

    [Serializable()]
    public class Supervisor
    {
        public string name = "";

        public Supervisor(string eName)
        {
            name = eName;
        }
    }
}

Edit: Edited based on jdweng's comment. I think jdweng is right.