1 Answers
int x = 3000, y = 2000 ;
long int z = x * y ;
Here the multiplication is carried out between two ints x and y, and the result that would
overflow would be truncated before being assigned to the variable z of type long int.
However, to get the correct output, we should use an explicit cast to force long arithmetic as
shown below:
long int z = ( long int ) x * y ;
Note that ( long int )( x * y ) would not give the desired effect.