[Interface2NullObject]Debugger and Converter–part 5

It will be interesting to see the Null Object in the debugger with properties generated from interface -see  https://www.nuget.org/packages/rscg_Interface_to_null_object

This is not very difficult to generate once you have the interface with properties

The code for generating

public string DebuggerDisplay()  
{
    StringBuilder sb = new StringBuilder("[System.Diagnostics.DebuggerDisplay(");
    sb.Append("\"");
    foreach (var item in props)
    {
        sb.Append(item.Name);
        sb.Append(" = {");
        sb.Append(item.Name);
        sb.Append("} " );
    }
    sb.Append("\")]");
    return sb.ToString();
}

and the code generated is

[System.Diagnostics.DebuggerDisplay("FirstName = {FirstName} LastName = {LastName} Department = {Department} ")]
public partial class Employee_null : global::IntegrationConsole.IEmployee

More interesting is to have a converter in order to obtain from the JSON string the interface,not the object .

The code for generating is

public class @(Model.Name + "Converter") : System.Text.Json.Serialization.JsonConverter<@(Model.FullName)>
    {
    public override @Model.FullName Read(ref System.Text.Json.Utf8JsonReader reader,System.Type typeToConvert,System.Text.Json.JsonSerializerOptions options)
    {
    // Deserialize the JSON to the concrete type @Model.Name
    return System.Text.Json.JsonSerializer.Deserialize<@nameClass>
        (ref reader,options);
        }

        public override void Write(System.Text.Json.Utf8JsonWriter writer,@Model.FullName value,System.Text.Json.JsonSerializerOptions options)
        {
        // Serialize the concrete type @Model.Name
        System.Text.Json.JsonSerializer.Serialize(writer,(@nameClass)value,options);
        }
        }

and the code generated is

//serialize and deserialize
var empString = JsonSerializer.Serialize(employee);
Console.WriteLine(empString);
//deserialize

var options = new JsonSerializerOptions
{
    PropertyNameCaseInsensitive = true,
    DefaultBufferSize = 128
};
options.Converters.Add(new IDepartmentConverter());
options.Converters.Add(new IEmployeeConverter());

var emp2 = JsonSerializer.Deserialize<IEmployee>(empString,options);
ArgumentNullException.ThrowIfNull(emp2);
Console.WriteLine(emp2.FirstName);
Console.WriteLine(emp2.Department.Name);
Debug.Assert(emp2.FirstName == "Andrei");