Why is list when passed without ref to a function acting like passed with ref?

Question

If I did not get this terribly wrong, this behaviour is strange for me. Rather than explaining, I'll post a sample code below and please tell me why does I get output x and not y.

    private void button1_Click(object sender, EventArgs e)
    {
        List<int> l = new List<int>() { 1, 2, 3 };
        Fuss(l);
        MessageBox.Show(l.Count.ToString());
    }

    private void Fuss(List<int> l)
    {
        l.Add(4);
        l.Add(5);
    }

Output should, I assume would be 3. But I get the output as 5. I understand the output can be 5 if I do this:

    private void button1_Click(object sender, EventArgs e)
    {
        List<int> l = new List<int>() { 1, 2, 3 };
        Fuss(ref l);
        MessageBox.Show(l.Count.ToString());
    }

    private void Fuss(ref List<int> l)
    {
        l.Add(4);
        l.Add(5);
    }

Answer 1

Parameters are passed by value in C# unless they are marked with the ref or out modifiers. For reference types, this means that the reference is passed by value. Therefore, in Fuss, l is referring to the same instance of List<int> as its caller. Therefore, any modifications to this instance of List<int> will be seen by the caller.

Now, if you mark the parameter l with ref or out, then the parameter is passed by reference. What this means is that in Fuss, l is an alias for storage location used as a parameter to invoke the method. To be clear:

public void Fuss(ref List<int> l)

called by

List<int> list = new List<int> { 1, 2, 3 };
Fuss(list);

Now, in Fuss, l is an alias for list. In particular, if you assign a new instance of List<int> to l, the caller will see that new instance assigned to the variable list as well. In particular, if you say

public void Fuss(ref List<int> l) {
    l = new List<int> { 1 };
}

then the caller will now see a list with one element. But if you say

public void Fuss(List<int> l) {
    l = new List<int> { 1 };
}

and call by

List<int> list = new List<int> { 1, 2, 3 };
Fuss(list);

then the caller will still see list as having three elements.

Clear?

Answer 2

It does not act like its passed by ref.

void ChangeMe(List<int> list) {
  list = new List<int>();
  list.Add(10);
}
void ChangeMeReally(ref List<int> list) {
  list = new List<int>();
  list.Add(10);
}

Try it. Do you notice the difference?

You can only change the contents of list (or any reference type) if you pass it without a ref (because as others have said, you are passing a reference to the object on the heap and thus change the same "memory").

However you cannot change "list", "list" is a variable that points to an object of type List. You can only change "list" if you pass it by reference (to make it point somewhere else). You get a copy of the reference, which if changed, can only be observed inside your method.