Complex property types

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I have a class named TerminalConnectionParameters which contains a few differently typed fields, like Hostname, IPAddress, UserName, Password, etc.

I want to used this class type as a property of a Dummy class, like :

C#

public class Dummy
{
    private TerminalConnectionParameters _parameters;
    public TerminalConnectionParameters Parameters
    {
      get
      {
        // do some fancy calculation here
        return _parameters;
      }
      set
      {
        _parameters = value;
        // do some fancy calculation here
      }
    }
}

This ensures that I can get or set the whole Parameters class of a Dummy object.
So far so good, but what, if I want to use this property like :

C#

objectDummy.Parameters.Hostname= "dummyRouter";

For sure, this assignment statement will not set the value of the dummyObject's internal _parameters field. But what then ? And what is the correct way to implenment such property writer without having to create a property accessor of each field of the TerminalConnectionParameters uniquely ?
I am aware of the trivial way as below, but is there no better solution ?

C#

TerminalConnectionParameters  temp = objectDummy.Parameters;
temp.HostName = "dummyRouter";
objectDummy.Parameters = temp;

And the reason behind this question is that I want to use PropertyGrid to deal with the internal _parameters object of a Form.

What I have tried:

As outlined in the example above.

Posted 20-Oct-16 2:34am

Laszlo Frank

Updated 20-Oct-16 9:46am

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3 solutions

Solution 1

You could serialize the parameters in the dummy object, and deserialize it in the host object.

Posted 20-Oct-16 3:51am

#realJSOP

Solution 3

Thank you Midi_Mick, that was it. Somehow I felt it must be simple...:-)

Posted 20-Oct-16 9:46am

Laszlo Frank

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Midi_Mick · Accepted Answer · 2016-10-20T04:24:00

Actually, your first assignment statement is good providing your _parameters member has been initialised - you can either do that in the declaration

C#

private TerminalConnectionParameters _parameters = new TerminalConnectionParameters();

or in the getter

C#

private TerminalConnectionParameters _parameters = null;
    public TerminalConnectionParameters Parameters
    {
      get
      {
        if (_parameters - null)
            _parameters = new TerminalConnectionParameters();
        // do some fancy calculation here
        return _parameters;

Now, once you are ensured of initialisation, all you need to do is decorate the Parameters property with the following attribute, and the designer will magically take care of the rest:

C#

[DesignerSerializationVisibility(DesignerSerializationVisibility.Content)]
public TerminalConnectionParameters Parameters
{