Boxing vs Unboxing

Question

Another recent C# interview question I had was if I knew what Boxing and Unboxing is. I explained that value types are on Stack and reference types on Heap. When a value is cast to a reference type, we call it boxing and vice versa.

Then he asked me to calculate this:

int i = 20; object j = i; j = 50; 

What is i?

I messed it up and said 50, where its actually 20. Now I think understand it why, however when I was playing with different combinations I was surprised to see this:

Object a = 1; // Boxing Object b = a; // referencing the pointer on stack to both objects on heap a = 2; // Boxing 

I was expecting to see b == 2 as well, but it isn't, why? Is it because the second boxing destroys and replaces the whole a object on the heap?

Because if I do this, it's fine:

public class TT {     public int x; }  TT t = new TT(); t.x = 1; TT t2 = new TT(); t2.x = 2; t = t2; t.x = 3; 

What is t2.x? It should be 3, and it is. But this is not an example of boxing / unboxing at all, is this correct? So how would you summarize this?

Could the values ever become the same in a boxing/unboxing conversion as above?

Answer 1

Very short: boxing means creating a new instance of a reference type. If you know this, you understand that one instance does not change by creating another.

What you are doing with a = 2 is not changing the value in the 'box', you are creating a new instance of a reference type. So why should anything else change?