own exception class hierarchy compile error
i was trying following program
package catchingexceptionhierarchy
class MyParentException extends Exception {
}
class MyChildException extends MyParentException {
}
public class Main {
public static void main(String[] args) {
try {
throw new MyChildException();
} catch (MyParentException s) {
System.err.println("Caught
} catch (MyChildException a) { // Compile error expected
System.err.println("Caught
}
}
}
i was not clear why 'Caught MyParentException' not called. since MyChildException extends MyParentException. How compil;e time error different from earlier progarm ion previous question without compiler error please advise
package catchingexceptionhierarchy
class MyParentException extends Exception {
}
class MyChildException extends MyParentException {
}
public class Main {
public static void main(String[] args) {
try {
throw new MyChildException();
} catch (MyParentException s) {
System.err.println("Caught
} catch (MyChildException a) { // Compile error expected
System.err.println("Caught
}
}
}
i was not clear why 'Caught MyParentException' not called. since MyChildException extends MyParentException. How compil;e time error different from earlier progarm ion previous question without compiler error please advise
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CEHJ
will have already been called by the first block=will have already been caught by the first block
ASKER
>>>The compiler knows that MyChildException will have already been called by the first block, since the first block that's type-correct does the business.
>>will have already been called by the first block=will have already been caught by the first block
what you mean by first block. you mean first catch block?
>>>
public static void main(String[] args) {
try {
throw new MyChildException();
} catch (MyParentException s) {
System.err.println("Caught
} catch (MyChildException a) { // Compile error expected
System.err.println("Caught
}
}
}
i thought MyChildException in second block is called here. please advise
>>will have already been called by the first block=will have already been caught by the first block
what you mean by first block. you mean first catch block?
>>>
public static void main(String[] args) {
try {
throw new MyChildException();
} catch (MyParentException s) {
System.err.println("Caught
} catch (MyChildException a) { // Compile error expected
System.err.println("Caught
}
}
}
i thought MyChildException in second block is called here. please advise
ASKER
>>The compiler knows that MyChildException will have already been called by the first block, since the first block that's type-correct does the business
can you please let me what do you mean by first block. where first bloc is calling MyChildException in this program . i am still not clear
can you please let me what do you mean by first block. where first bloc is calling MyChildException in this program . i am still not clear
The first block is
>>
} catch (MyParentException s) {
System.err.println("Caught
>>
which will catch MyChildException
>>
} catch (MyParentException s) {
System.err.println("Caught
>>
which will catch MyChildException