Question

As title, just after return statement?

int x = 1;

function F(ref int y) { y  = y + AnotherF(x); }

function AnotherF(result z)
{
   z = null;
   return (-1);
}

F(x); print(x); // prints 0 or null?

Answer 1

Answer to myself: The function AnotherF returns -1. The just before destroying its record (and thus before passing the control back to F), the value of z is assigned back to the actual parameter (y). After assigning null to x then F continues to evaluate y = 1 + (-1) = 0. Then x is 0.