I was practicing exception handling mechanisms with method overriding in java...My code is as follows:
class base {
void show() {
System.out.println("in base class method");
}
}
class derived extends base {
void show() throws IOException {
System.out.println("in derived class method");
throw new IOException();
}
}
class my {
public static void main(String[] args) {
try {
base b = new derived();
b.show();
}
catch (IOException e) {
System.out.println("exception occurred at :" + e);
}
}
}
Showing an error :
So, I corrected following:
void show() throws IOException{
and it is working correctly...
One more experiment I did :
void show() throws Exception{
but it was also showing an error:
As I understand, this is because of an overridden method's throws
clause should mention the exact checked exception in throws
clause of super class method.
As in second case, if I write IOException
's superclass Exception
in throws clause, it also shows an error. why? Even if Exception
is parent class of all exceptions.
I just experimented...what this error tells I don't know...
Can any one please explain what it says and what are the constraints for mentioning checked exception in throws
clause of an overridden method?
There are two related errors in the sample:
1) Your base class method provides the "template" or basic criteria for the derived class method.
So, the base class should declare a super-set ie either the same exception class or base exception class of the derived class. You cannot declare that it throws nothing, because then the criteria will not match.
So if your derived class method is like this:
class Derived extends Base {
void show() throws IOException {
//...
}
}
Then base class method "must" be:
class Base {
void show() throws /*Same or base classes of IOException*/ {
//...
}
}
So both of these work:
class Base {
void show() throws Exception {
//...
}
}
or
class Base {
void show() throws Throwable {
//...
}
}
2) When you try the above, the overall declaration of your show
method now becomes throws Exception
. As a result, anyone who uses this show
must catch that exception.
In your main
method, you are catching IOException
. This will no longer work, the compiler complains "ok you are catching the IOException, what about all the other possibilities from Exception?" This is the second error that you showed.
To fix this, change the main
method catch to include Exception
as declared in the base class:
class My {
public static void main(String[] args) {
try {
base b = new derived();
b.show();
}
/* NOTE: CHANGED FROM IOException TO Exception */
catch (Exception e) {
System.out.println("exception occurred at :" + e);
}
}
}
Overridden methods may contain only the same checked exceptions in their throws
clause as the super-method, or derived types at most.
For example if you say
class Base {
public void foo(int y) throws IOException {
// ...
}
}
and
class Derived extends Base {
public void foo(int y) throws Exception {
// ...
}
}
then your compiler will say that the foo
method inside Derived
is incompatible with the throws clause in its superclass.
The other way around works because if I say
class Base {
public void foo(int y) throws Exception {
// ...
}
}
and
class Derived extends Base {
public void foo(int y) throws IOException {
// ...
}
}
it's OK.
Why.
Think about the usage of your methods. Java expects you to be using the method polymorphically such as
Base a = new Derived();
a.foo(3);
As such, the compiler will force you to catch the exception thrown by foo
in your declared type of the variable ( Base
). So your code will become
Base a = new Derived();
try {
a.foo(3);
} catch (Exception e) {
// ...
}
Therefore, the subtype of Exception
you declared in the Derived
type is OK with your code above (a catch for an Exception
will work for any of its subtypes as well) and as such, Java will allow you to declare IOException
in derived, because it will cause no worries later on.
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.