Can anyone please give me information about it. I cant really uderstand the type of this Exception.
Thank you
public class ValidationException extends Exception{
public ValidationException(){
super("There was a problem when validating data");
}
public ValidationException(String message){
super(message);
}
public ValidationException(String message, Throwable throwable){
super(message, throwable);
}
public ValidationException(Throwable throwable){
super(throwable);
}
}
It is a "runtime exception" in the (fatuous) sense that it is an exception that occurs at runtime. But that is true for all Java exceptions ... apart from bugs in the compiler, etcetera.
It is not a subclass of RuntimeException. You have declared it as a subclass of Exception and Exception is not a subclass of RuntimeException. (In fact, the reverse is true: RuntimeException is a subclass of Exception!)
It is a checked exception because it is not a subclass of RuntimeException (or Error).
Since it is a checked exception, the Java rules about checked exceptions apply. For example, any method that throws or propagates1 your exception must declare that it throws this exception, or an exception that is a superclass of this exception.
1 - Technically, the JLS describes this as an abnormal termination of the method body with this exception as the abnormal termination reason.
The main difference between Exception, and RuntimeException is that we need to wrap a Exception in a try/catch block. A RuntimeException does not need to be caught, but it is just as lethal as an Exception.
public class Main{
public static void main(String[] args) {
Thread.currentThread().setUncaughtExceptionHandler(
new Thread.UncaughtExceptionHandler(){
#Override
public void uncaughtException(Thread t, Throwable e){
System.out.println("Uncaught Exception " + e);
}
});
try{
throwException();
}catch(Exception e){
System.out.println("Caught Exception " + e);
}
try{
throwRuntimeException();
}catch(Exception e){
System.out.println("Caught RuntimeException " + e);
}
//unchecked, no need to wrap int try/catch
throwRuntimeException();
}
public static void throwException() throws Exception {
throw new Exception();
}
public static void throwRuntimeException() {
throw new RuntimeException();
}
}
Take this example above. The output is this:
Caught Exception java.lang.Exception
Caught RuntimeException java.lang.RuntimeException
Uncaught Exception java.lang.RuntimeException
As you can tell, the call to throwRuntimeException() gets thrown, and since there is no try/catch block it has no idea how to handle it. This crashes the thread and since there is an UncaughtExceptionHandler it gets called.
Then there is also Error which I won't go into since I don't know much about it besides that JVM throws it. OutOfMemoryError is an example.
Related
#Override
Public class example {
void test {
try {
someMethod(); //This throws TimeoutException
} catch (TimeoutException ex) {
throw new TimeoutException(ex); //It doesn't throw error if I replace this with throw new RuntimeException(ex)
} }
}
The above example gives an error as 'throw new TimeoutException(ex)' as "TimeoutException(java.lang.string) in java.util.concurrent.TimeoutException cannot be applied to (java.util.concurrent.TimeoutException)".
But it doesn't throw an error if I replace it with 'throw new RuntimeException(ex)';
TimeoutException doesn't have a constructor that accepts a TimeoutException as an argument of the form TimeoutException(TimeoutException cause) or similar.
You can instead:
TimeoutException localtoe=new TimeoutException("test failed");
localtoe.initCause(ex);
throw localtoe;
Or equally:
throw new TimeoutException("test failed").initCause(ex);
initCause() may only be called once and only if cause wasn't set by a constructor. It's a funny little method that acts like a constructor after-thought(*).
There's nothing necessarily wrong with wrapping an exception as the cause of an exception.
Suppose testFunction() connects and then performs some operation.
You might want to throw an exception with message "connection failed in testFunction" and another "operation failed in testFunction" depending on what sub-step failed.
But if you don't need to provide so much detail you can just throw ex or let the method unwind without itself catching anything.
Here's a little example:
import java.util.concurrent.TimeoutException;
class Example{
private static void connect() throws TimeoutException {
//Dummy connection that just fails...
throw new TimeoutException("connection failed");
}
private static void process() throws TimeoutException {
try {
connect();
}catch(TimeoutException toe){
TimeoutException toeout=new TimeoutException("process failed because connection failed.");
toeout.initCause(toe);
throw toeout;
}
//Code for when connection succeeds...
}
public static void main (String[] args) throws java.lang.Exception
{
try{
process();
}catch(TimeoutException toe){
System.out.println(toe);
}
}
}
Expected output:
java.util.concurrent.TimeoutException: process failed because connection failed.
(*) initCause() looks like an after-thought and is somewhat. It was added to Java 1.4 in 2002. The documentation talks about 'legacy' constructors. Rather than double up the number of constuctors (to add one with a Throwable cause argument) it appears it was decided to allow this as bolt-on initialization.
It's debatable whether that was the best solution.
Things I observed in your question
you are trying to call a method directly from class in a try cache block. which is wrong you have to create method and call it from that
you want to throw an exception. SO you have to throw it method level from where you are calling it
please find the demo solution below :
public class example {
public void testFunction() throws TimeoutException {
try {
someFunction();
} catch (TimeoutException ex) {
throw ex;
}
}
public void someFunction() throws TimeoutException {
}
}
Java has 2 categories of exceptions: checked and unchecked. Checked exception (usually subclasses of Exception) must be declared in function signatures, while unchecked ones (usually subclasses of RuntimeException) must not.
TimeoutException is a checked exception. When it could be thrown from a method that does not declare it, you have 2 options:
declare it in the signature:
public void func1() throws TimeoutException {
somefunction();
}
clean and simple but it can be problemetic is func1 is an override of a function not declared to throw this exception, of it it is called from another function (suppose from a framework) that does not declare it either
hide it in an unchecked exception
public void func1() {
try {
somefunction();
} catch (TimeoutException e) {
throw new RuntimeException(e);
}
}
you lose the declarative part (checked exceptions exist for that reason), but at least it allows you to call it from function not declaring it.
You have roughly three options here:
Rethrow the same exception: `throw ex;'
Throw a new TimeoutException and lose the stack trace: throw new TimeoutException(ex.getMessage());
Throw an exception of another type, such as RuntimeException.
Each of these options have advantages and drawbacks, you decide.
UPDATE (thanks to #Mark Rottenveel)
Point 2 could be rewritten: throw new TimeoutException(ex.getMessage()).initCause(ex); to keep the link to the original exception.
I have these two java codes:
class test {
public static void main(String[] args) throws IOException {
System.out.println("Before try”");
try{}
catch(Throwable d ) {}
System.out.println("done");
}
}
it will compile and print Before try and done.
class test {
public static void main(String[] args) throws IOException {
System.out.println("Before try”");
try{}
catch(java.io.IOException e ) {}
System.out.println("done");
}
}
}
this will make compilation error:
exception java.io.IOException is never thrown in body of corresponding
try statement
at javaapplication8.test.main(Try.java:63)
what is the difference between throwable exception and IOException to have these results, Is there a rule to know which exception would need to be thrown or not?
Java has a concept of checked exceptions. All Throwable are checked unless they are a sub-class of Error or RuntimeException
In the catch class, the compiler can work out whether you can throw a checked exception or not (under normal conditions), but it can't tell if you have thrown an unchecked exception so if you catch any unchecked exception, or an parent, it cannot determine whether you might throw it or not.
Your declaration of main promised that it could throw an IOException, but your code broke that promise (because you catch it yourself).
I have a method, in which right now I am handling the exception using the try catch method.
I have a custom exception method to handle the error.
Now I have to change this exception handling to runtime exception.
Code:
public class AppException extends RuntimeException {
private static final long serialVersionUID = -8674749112864599715L;
public AppException() {
}
public AppException(String message, Throwable cause,
boolean enableSuppression, boolean writableStackTrace) {
super(message, cause, enableSuppression, writableStackTrace);
}
public AppException(String message, Throwable cause) {
super(message, cause);
}
public AppException(String message) {
super(message);
}
public AppException(Throwable cause) {
super(cause);
}
}
Method which handled using try catch.
#Transactional(readOnly = false)
#Override
public String save(StagingDocument stagingData)
throws AppException {
String enrichObjectId = null;
try {
EnrichDocument document = getEnrichDocument(stagingData);
EnrichDocument enrichPayload = enrichStagingDocumentRepository
.save(document);
enrichObjectId = enrichPayload.getId().toString();
} catch (Exception e) {
logger.error("EXCEPTION IN SAVETOENRICHDOCUMENT METHOD: " + e);
throw new AppException (e.getMessage(), e.getCause());
}
return enrichObjectId;
}
Here in the above method is the implementation when AppException extends the Exception class.
Now I need to change the save method according to runtime exception handling.
Question:
How can I change this method,without using try catch method?
If try catch is not there how the exception is get handled?
RuntimeExceptions don't have to be declared in the method signature, and if you want to remove the try/catch blocks as well - you can do it:
#Transactional(readOnly = false)
#Override
public String save(StagingDocument stagingData) {
String enrichObjectId = null;
EnrichDocument document = getEnrichDocument(stagingData);
EnrichDocument enrichPayload = enrichStagingDocumentRepository
.save(document);
enrichObjectId = enrichPayload.getId().toString();
return enrichObjectId;
}
When there's no try/catch - the exception doesn't get "handled" it is rather getting cascaded until either a higher level handles it or until the highest level exists the program (with the RuntimeException).
If you don't use a try-catch block, the original Exception would not be caught. If it is a RuntimeException, you will get no compilation errors as RuntimeExeption does not need to be caught. If that Exception occurs, it will simply be thrown from the save() method (it will be delegated).
If the original Exception is not a RuntimeException and you don't want to use a try-catch block, you may declare the save() method to throw that Exception but in this case it will obviously be not an instance of AppException but the original exception itself.
Btw, if you create an AppException, its cause should be e and not e.getCause(). If you pass e.getCause() as the cause of your AppException, then e itself will be lost. You might also want to add a custom error message and not use the original exception's message.
Summary: if you want your save() method to throw an instance of AppException if an Exception is encountered inside, you can't do that without a try-catch block, you have to catch it (the Exception encountered inside) and wrap it in a new AppException like you did.
Runtime exceptions need not be caught. If you are using AppException as a checked exception, still you need not put try/catch since you have already handled it using throws in the method signature. The method which is calling save() will have to handle the AppException.
I get this error when compiling my program:
./GUI/mainWindow.java:30: error: unreported exception Exception; must be caught or declared to be thrown
clientUI.initClientUI();
^
1 error
But my method actually throws Exception:
public final void initClientUI() throws Exception {
However, if I remove the "throws Exception" part, try to compile, then add it again, the program compiles successfully. Can anyone explain why is this happening?
EDIT:
If I add "throws Exception" to the method I am calling
public void actionPerformed(ActionEvent e) throws Exception { //<-----add Exception there
ClientWindow clientUI = new ClientWindow();
clientUI.initClientUI();
I get this error:
./GUI/mainWindow.java:28: error: actionPerformed(ActionEvent) in <anonymous GUI.mainWindow$1> cannot implement actionPerformed(ActionEvent) in ActionListener
public void actionPerformed(ActionEvent e) throws Exception {
^
overridden method does not throw Exception
1 error
By saying
public final void initClientUI() throws Exception {
you are telling the compiler that initClientUI() is allowed to throw Exception or any subclass of Exception. Since Exception is checked, any method that calls initClientUI() must either catch Exception or have it listed in its throws clause.
I'd recommend reading the tutorial on exceptions.
Your problem is not the declaration of your method, your problem is when you call it. When you declare a method like this
public final void initClientUI() throws Exception
your compiler will know that it will throw an exception. That's good, no problem so far. But when you call it, you have to deal with this exception.
So when you say
client.initClientUI()
your compiler says "this method throws an exception, pleas handle it". Now you have two options:
try {
client.initClientUI()
}
catch(Exception e) {
System.out.println(e);
}
or declare your method that calls client.initClientUI() also with throws Exception.
throws Exception means you are throwing it to the calling method,And Calling method should be responsible of handeling that exception either by throwing it again or catching it with try catch.
You should do it this way
public void actionPerformed(ActionEvent e) {
try{
ClientWindow clientUI = new ClientWindow();
clientUI.initClientUI();
}catch(Exception e){
// handle exception
}
I have the following code. In that fn2 is actually throwing an exception and it is caught in the function itself. In the function fn1 the compiler is complaining about unhandled exception since fn2 is declared to be throwing an Exception.
Why it is so? Since the exception is caught inside fn2 it shouldn't be complaining right?
Please explain the behavior.
public class ExepTest {
/**
* #param args
*/
public static void main(String[] args) {
ExepTest exT = new ExepTest();
exT.fn1();
}
public void fn1(){
fn2();//compilation error
}
public void fn2() throws Exception{
try{
throw new Exception();
}
catch(Exception ex){
System.out.println("Exception caught");
}
}
}
Compiler doesn't/can't know that at runtime no exception will be throwed by fn2() since it's declared that it may throw an Exception, that's why you got an error.
Remove the throws Exception from fn2() signature
The signature of the method fn2 is all that matters here. In that signature you declare that fn2 may throw an Exception. Any code that calls a method that may throw an Exception must handle the eexception.
public void fn2() throws Exception. The compiler see that declaration and expect each caller to fn2 to handle / rethrow the exception.
Exception is thrown BY fn2, not inside it. So it will be actually thrown where it is called. Since it is called in fn1, it is behaving like this.
You need to surround the call to fn2() in a try-catch block, or declare that fn1 also throws an exception.
public void fn1(){
fn2();//compilation error
}
public void fn2() throws Exception{
try{
throw new Exception();
}
catch(Exception ex){
System.out.println("Exception caught");
}
}
}
Here compiler will not recognize that you have handled exception or not. It just assumes that fn2 throws exception as you have declared to be and that's why it's showing error.
To run program either remove throws Exception from fn2 or write throws Exception in fn1 or handle it in try..catch in fn1.