>An object is passed by reference. I believe that this is due to passing the memory address of the object, not the physical object (or something along those lines).

Right. You are actually passing a memory pointer to the function. The function then manipulates the value in that address. When you RETURN out of the function, the calling routine's variable has been altered because the data in its address has been manipulated.

Of course, I am assuming there is some manipulation...

Joe