c# convert existing class to use properties correctly

Question

I have the following classes:

class Given
{
    public string text = "";
    public List<StartCondition> start_conditions = new List<StartCondition>();
};

class StartCondition
{
    public int index = 0;
    public string device = "unknown";
    public string state = "disconnected";
    public bool isPass = false;
};

I want to convert them into c# properties (using get; and set;)

Looking at this question: what-is-the-get-set-syntax-in-c, it seems I can make a property nice and easy like this:

class Given
{
    public string text { get; set; }
    public List<StartCondition> start_conditions { get; set; }
};

class StartCondition
{
    public int index { get; set; }
    public string device { get; set; }
    public string state { get; set; }
    public bool isPass { get; set; }
};

But now I don't know how I should add my initialisations, because I want the same start values as I had before, or for the List container I want it to be new'ed.

What is the best way to achieve this?

Are you on c# 6? Because if so, just add `{ get; set; } = 0`, etc. Otherwise, just initialize in the constructor. — Kirk Woll, Nov 07 '16 at 17:45
You can also add constructors to the classes to and make them set those parameters any time the class is instantiated. — B.Rickman, Nov 07 '16 at 17:45
@KirkWoll Thanks very much - seems obvious now :o ... I have C#2015, not sure what C# 6 is? — code_fodder, Nov 07 '16 at 17:57
@B.Rickman Also a good point!, for some reason I was not sure if the name of the "private" (hidden?) variable is the same as the "public" one — code_fodder, Nov 07 '16 at 17:59
@code_fodder, if you're on VS2015, then you have access to C# 6 features. — Kirk Woll, Nov 07 '16 at 18:26
@KirkWoll, ah ok thanks!... I guess this is a standards version, like C++11 / C++14? — code_fodder, Nov 07 '16 at 18:28
Yeah, basically. Typically new versions of VS come with a new version of C#, though they are not strictly correlated. — Kirk Woll, Nov 07 '16 at 18:28

score 6 · Accepted Answer · answered Nov 07 '16 at 17:46

6

The ability to have auto property initializers is included since C# 6.0. The syntax is:

public int X { get; set; } = x; // C# 6 or higher

answered Nov 07 '16 at 17:46

Yoshiaki Nakamura

233
2
8

Thanks very much. Is C# 2015 > C#6? ... I assume it is – code_fodder Nov 07 '16 at 18:00

score 4 · Answer 2 · answered Nov 07 '16 at 17:45

4

Use a constructor. So your class would look like this:

public class StartCondition
{
    public int index { get; set; }
    public string device { get; set; }
    public string state { get; set; }
    public bool isPass { get; set; }

   // This is the constructor - notice the method name is the same as your class name
   public StartCondition(){
      // initialize everything here
      index = 0;
      device = "unknown";
      state = "disconnected";
      isPass = false;
  }
}

answered Nov 07 '16 at 17:45

Darren Wainwright

30,247
21
76
127

This is what i would do, as i said in the comments of the question. =] – B.Rickman Nov 07 '16 at 17:47
Same here - and yes, i was typing as you added a comment. Feel free to add an answer if you prefer..... – Darren Wainwright Nov 07 '16 at 17:47
Nope, you beat me to it! i upvoted your answer as i think its the best. – B.Rickman Nov 07 '16 at 17:48

score 0 · Answer 3 · answered Nov 07 '16 at 17:51

Create a Constructor to start your class instance with the default values

class Given
{
    public Given(){
        this.text = "";
        start_conditions = new List<StartCondition>();
    }
    public string text { get; set; }
    public List<StartCondition> start_conditions { get; set; }
};

class StartCondition
{
    public StartCondition(){
        this.index = 0;
        this.device = "unknown";
        this.state = "disconnected";
        this.isPass = false;
    }
    public int index { get; set; }
    public string device { get; set; }
    public string state { get; set; }
    public bool isPass { get; set; }
};

Now you can create your instances with the default values by using StartCondition A = new StartCondition();

score 0 · Answer 4 · answered Nov 07 '16 at 17:52

If you are not using C# 6+ (or even if you are), you can explicitly declare your backing variables for properties:

public class Given
{
    private string _text = string.Empty;
    private List<StartCondition> _start_conditions = new List<StartCondition>();

    public string text { get{ return _text; } set{ _text = value; } }
    public List<StartCondition> start_conditions { get{ return _start_conditions; } set{ _start_conditions = value; } }
}

This allows you to set your initializations as before.

c# convert existing class to use properties correctly

4 Answers4

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