What will be the output of the program? try
{
Float f1 = new Float("3.0");
int x = f1.intValue();
byte b = f1.byteValue();
double d = f1.doubleValue();
System.out.println(x + b + d);
}
catch (NumberFormatException e) /* Line 9 */
{
System.out.println("bad number"); /* Line 11 */
}
Correct Answer: 9.0
Explanation:
The xxxValue() methods convert any numeric wrapper object's value to any primitive type. When narrowing is necessary, significant bits are dropped and the results are difficult to calculate.
