You can add any two doubles using the + operator.
~void main() {
double x = 5.1;
// y will be 14.2
double y = x + 9.1;
IO.println(y);
~}Because of the previously mentioned inaccuracy, the results of additions might not always be what you expect.
~void main() {
~double x = 5.1;
~// y will be 14.2
~double y = x + 9.1;
// z will be 19.299999999999997
double z = x + y;
IO.println(z);
~}You can add any int to a double and the result of any such addition will also be a double.
~void main() {
int x = 5;
double y = 4.4;
// z will be 9.4
double z = x + y;
IO.println(x);
IO.println(y);
IO.println(z);
~}Even if the result of such an expression will not have any fractional parts, you cannot directly assign it to an int.
~void main() {
int x = 5;
double y = 4;
// even though z would be 9, which can be stored in an int
// this will not work. The result of the expression is a double.
int z = x + y;
IO.println(x);
IO.println(y);
IO.println(z);
~}