Question

I have a class called Config with the declaration
public class Config<T> where T : class, new()

It contains some methods to save a configuration of type T as well as the actual configuration in property called Configuration.

Is it possible to expose the configuration of type T directly in my Config class without having to go through the Configuration property.

So far, I have a bit of an ugly work around

public T this[int index]
{
    get { return _configuration; }
}

If not, then I was wondering if it can be done by inheriting from the T, but I’m not clear on the syntax of how to do this.

Edit:
I imagined you could do something like

public T this
{
    get { return _configuration; }
}

Answer 1

You can use an implicit operator:

class Config<T> where T: class, new()
{
  private T _configuration;

  public static implicit operator T(Config cfg)
  {
    return cfg._configuration;
  }
}

Use it like so:

var config = new Config<SomeClass>();
SomeClass realConfig = config;