Question

i have a class

    class ABC
    {
          public int a = 2;
          public void valueA(ABC objabc)
          {
               a = 5;
               objabc = new ABC();
               objabc.a = 11;
           }
    }

then i write main as

    static void Main(string[] args)
    {
        ABC objabc = new ABC();
        objabc.a = 15;
        objabc.valueA(objabc);
        Console.WriteLine(objabc.a);
     }

When i execute this i found 5 in output.so my question is that why a = 5??why it is not 2, 11 or 15??

Answer 1

You are calling the valueA method on the object objabc. Inside the method the member a of this is set to 5. You can see the code as equivalent to:

class ABC
{
      public int a = 2;
      public void valueA(ABC objabc)
      {
           this.a = 5;
           objabc = new ABC();
           objabc.a = 11;
       }
}

So you are setting the member of the object on which you called the method to 5. This object being objabc from main, so the final value is 5. The fact that you are assigning it a new object reference afterwards makes no difference, because you are not passing it by reference so changes are not visible outside.

The result would be different (i.e. 11) if your code was this:

class ABC
{
      public int a = 2;
      public void valueA(ref ABC objabc)
      {
           a = 5;
           objabc = new ABC();
           objabc.a = 11;
       }
}

In this case the parameter would be passed by reference, so the assignment objabc = new ABC(); would be visible to the calling code (i.e. in Main).