Why must this method invocation be caught?!

Question

Hi there. I have this following piece of code:
public class Dog extends Animal {
     public static void main(String[] args) {
          Animal animal = new Dog();
          System.out.println(animal.getAge(args[0]));
     int getAge(String arg) {
          return Integer.parseInt(arg);
class Animal {
     int getAge(String x) throws Exception {
          return 5;
}Ok, I compiled this an received this notification from the compiler:
Dog.java:4: unreported exception java.lang.Exception; must be caught or declared
to be thrown
                System.out.println(animal.getAge(args[0]));
                                                ^
1 errorIm a little confused here as to why I am receiving this notification...
The reason being is firstly I thought that when you override a method you can declare fewer or narrower checked
exceptions than the method that is being overridden and secondly,
this is a polymorphic method invocation on line 4 as Dog's getAge method is being called,
not class Animal's getAge method.
Any clarification would be well received,
thank you indeed.
Edited by: shamrock08 on May 6, 2008 10:23 AM
Edited by: shamrock08 on May 6, 2008 10:23 AM

Answer 1

shamrock08 wrote:
BigDaddyLoveHandles wrote:
you don't get that compile-time error. Why do you get it with the first code? In expression animal.getAge(args[0]), variable animal is of type Animal, and Animal's getAge method signature includes "throws Exception". Typing is done statically, at compile-time, hence the syntax error.Im not sure im entirely understanding.
I thought that in relation to overridden instance methods, the Object type determines which method
will be invoked at runtime. Yes.
So in this case wont the Dog getAge method be invoked? Yes.
Im just a little confused as my book says that
when a method overrides an inherited instance method it can declare fewer or narrower checked exceptions,
Thanks for the input & regards.That's also true. I think what you are missing is the realization that compiler errors are based on static type information not runtime information.
Consider this method:
void f(Dog d, Animal a) {
    d.getAge("");
    a.getAge("");
}The method invocation d.getAge("") doesn't throw Exception, because that is how you overrode getAge in class Dog.
The method invocation a.getAge("") may throw a checked exception, because that is how the method is defined in class Animal. So the compiler will complain about this line: either catch or propagate the exception. Note that this compiler error is independent of what object a refers to -- a could be pointing to a Dog, it doesn't matter. Only the static typing matters at compile-time.