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

1. You’re right the second assignment replaces the first. It doesn’t change the boxed value.

2. Your example doesn’t make use of boxing. The value (int) is stored as an int and not boxed.

3. No, boxing still keeps the immutability guarantee.