CODEWITHSUNDEEP
c#
Georgy
Georgy

Reputation: 97

How to check whether there is within the class of the same class?

I'am sorry for my Engilsh. I have some class like this:

public class MainClass
{
      public string message { get; set; }
      public MainClass forward { get; set; }
}

And have Main funcion, where I'am inizialize class and fill data(in real project I have data in JSON format, where class can be embedded itself in an infinite number of times) of class:

static void Main(string[] args)
{
    MainClass clasClass = new MainClass()
    {
        message = "Test1",
        forward = new MainClass()
        {
            message = "Test1_1",
            forward = new MainClass() 
            {
                message = "Test1_1_1",
                forward = new MainClass()
                {
                    message = "Test1_1_1_1",
                    forward = new MainClass()
                }   
            }
        }
    };
}

How do I get the number of nested class names , without knowing their number?

Upvotes: 4

Views: 106

Answers (2)

M.kazem Akhgary
M.kazem Akhgary

Reputation: 19179

You can go forward and count!

int count = 0;
MainClass tempClass = clasClass;
while (tempClass.forward != null)
{
    count++;
    tempClass = tempClass.forward;
}

Also you can make this a bit smaller.

int count = 0;
MainClass tempClass = clasClass;
while ((tempClass = tempClass.forward) != null) count++;

Upvotes: 2

Jon Skeet
Jon Skeet

Reputation: 1503649

It sounds like you just want recursion:

public int GetNestingLevel(MainClass mc)
{
    return mc.forward == null ? 0 : GetNestingLevel(mc.forward) + 1;
}

Or as part of MainClass itself:

public int GetNestingLevel()
{
    return mc.forward == null ? 0 : mc.forward.GetNestingLevel() + 1;
}

Or in C# 6, if you want to use the null conditional operator:

public int GetNestingLevel()
{
    return (mc.forward?.GetNestingLevel() + 1) ?? 0;
}

That could cause problems if you have very deeply nested classes, in terms of blowing the stack - but it's probably the simplest approach otherwise. The alternative is to use iteration, as per M.kazem Akhgary's answer.

Upvotes: 4

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