Re: Defining a cast on a templated class

I'm creating a sort-of "wrapper" class which (partly) acts like a
variable. Something like:
template<class t>
class cell{
  t curval;
public:
  /*public for debugging only - will be private in final version*/
  inline cell<t>& set_value(t v){ curval = v; return *this;}
  inline t get_value(){ return curval;}
  /*actual public interface*/
  /*assign to this value*/
  inline cell<t>& operator=(t v){ return set_value(v);}
};
...
int
main(void){
  cell<int> v;
  v = 0;
  cout << "v = " << ((int)v);
  return 0;
}

However I can't figure out how to make the compiler do v.get_value()
when v is used in an int() context. Since this is templated, the only
conversion I want to support is something like cell<type> -> type.
Attempts to convert to unrelated types should signal an error.

I can't figure out the correct syntax to do this, unfortunately
(should probably look for better manuals, sigh).

----

In any case the actual target would be something like this:
int
main(void){
  cell<int> v;
  cell<int> v2;
  v = v2 + 1; //v automatically changes when v2 is changed
  v2 = 3;
  if((int)v == 4){
    cout << "Test 1 passed\n";
  } else{
    cout << "Test 1 failed\n";
    return 1;
  }
  v2 = 4;
  if((int)v == 5){
    cout << "Test 2 passed\n";
  } else{
    cout << "Test 2 failed\n";
    return 1;
  }
  return 0;
}

Basically the cell<> class would be like a spreadsheet cell - when I
assign a formula into it, its value will be automatically updated if
any of the cells in the formula are updated.

The way I intend to implement it is, I extend +-*/ and ?: so that if
any cells are in any parameters, they will instead return a new cell
which registers itself to the input cells. Something like:
template<class t>
cell<t>& operator+(cell<t> v1, t v2){
  sum_cell<t> *sum = new sum_cell<t>(v2);
  sum.register_yourself_to(v1); //when v1 is update()d, sum's update()
method is called
  return (cell<t>&) *sum;
}

If there is already some existing library that does (in some form) the
above, please inform me. I suspect there already is, somehow I have a
feeling I've seen it before.