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How to make sure that exceptions do not lose its stack trace information in java?

Recently, I met some exception problem in java, which reminded me of the typical idiom recommended by Bruce Eckel:
Converting checked to unchecked exceptions
The real problem is when you are writing an ordinary method body, and you call another method and realize, "I have no idea what to do with this exception here, but I don’t want to swallow it or print some banal message." With chained exceptions, a new and simple solution prevents itself. You simply "wrap" a checked exception inside a RuntimeException by passing it to the RuntimeException constructor, like this:
try {
// ... to do something useful
} catch (IDontKnowWhatToDoWithThisCheckedException e) {
throw new RuntimeException(e);
}
This seems to be an ideal solution if you want to "turn off the checked exception—you don’t swallow it, and you don’t have to put it in your method’s exception specification, but because of exception chaining you don’t lose any information from the original exception.
This technique provides the option to ignore the exception and let it bubble up the call stack without being required to write try-catch clauses and/or exception specifications.
However, I found it didn't work in some cases. as seen here:
package exceptions;
// How an exception can be lost
class VeryImportantException extends Exception {
#Override
public String toString() {
return "A very important exception";
}
}
class HoHumException extends Exception {
#Override
public String toString() {
return "A trivial exception";
}
}
public class LostMessage {
void f() throws VeryImportantException {
throw new VeryImportantException();
}
void dispose() throws HoHumException {
throw new HoHumException();
}
public static void main(String[] args) {
try {
LostMessage lm = new LostMessage();
try {
lm.f();
} catch (VeryImportantException e) {
throw new RuntimeException(e);
} finally {
lm.dispose();
}
} catch (Exception e) {
throw new RuntimeException(e);
}
}
}/* Output:
Exception in thread "main" java.lang.RuntimeException: A trivial exception
at exceptions.LostMessage.main(LostMessage.java:36)
Caused by: A trivial exception
at exceptions.LostMessage.dispose(LostMessage.java:23)
at exceptions.LostMessage.main(LostMessage.java:33)
*///:~
As the output demonstrated， the second exception completely obliterates the first one. There is no record of the first exception in the exception stack trace, which can greatly complicate debugging in real systems. usually, it’s the first exception that you want to see in order to diagnose the problem.
Joshua Bloch recommends the try-with-resource way that a resource must implement the AutoCloseable interface, which process is somewhat complex.
So, my question is this: is there any way I can use to make sure that exception will not lose its stack trace information by Bruce Eckel's approach?

You might want to consider using try-with-resource instead of a finally block. It tends to handle this situation more like it sounds you would want the situation handled. See this article for more details.
Alternatively, you could simply eat the exception (as Andy Thomas's answer shows), or (if you want to know about both exceptions that were thrown) you could combine the exceptions into a single kind of Aggregate Exception.

The problem isn't that you're wrapping the exception. The problem is that you're replacing it with a subsequent, unrelated exception thrown from the finally block.
One easy way to avoid this is to not throw an exception from the finally block.
For example:
try {
LostMessage lm = new LostMessage();
try {
lm.f();
} catch (VeryImportantException e) {
throw new RuntimeException(e);
} finally {
try {
lm.dispose();
} catch ( HoHumException e ) {
// No-op or logging
//
// If we're exiting this try-finally because an exception
// was thrown, then don't allow this new exception to replace it.
}
}
} catch (Exception e) {
throw new RuntimeException(e);
}

Am I not able to throw specific exception

#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.

Java: is there a semantical difference between declaring throws Exception and declaring which exceptions are thrown? [duplicate]

This question already has answers here:
In Java, is using throws Exception instead of throwing multiple specific exceptions good practice?
(15 answers)
Closed 6 years ago.
Final edit due to this question having been marked as duplicate: this question is about the semantics of the throws declaration - the question which is this said to be a duplicate of handles different aspects of throws and none of those 15 answers there gave me the insight of the chosen answer here. Anyway - so let's keep it here as a duplicate.
In Java you have to declare a throws clause and name the exceptions that this method could throw - so the easy way is to just declare
void myMethod(...) throws Exception
or you could be more specific and for example state
void myMethod(...) throws SQLException, NamingException
in case the method may just throw these two.
I understand that it makes a difference whether in a try/catch block I
try { ... } catch (Exception exc) { ... }
or
try { ... } catch (SQLException | NamingException exc) { ... }
because the first will also catch a RuntimeException, the second won't - so this is a difference.
But is there also any difference in declaring throws Exception vs. throws SQLException, NamingException concerning the semantics of the program? The second may be more readable, but I don't see any other difference. Is there any?

But is there also any difference in declaring throws Exception vs. throws SQLException, NamingException
The difference is:
if you declare throws SQLException, NamingException the compiler assures that you have catched exactly these two exceptions. Otherwise you will get a Unhandled exception type ... error.
On the other hand, if you declare throws Exception, the compiler assures that you have catched Exception.
In the second case, you can still catch any other exception which inherits from Exception without getting a compiler error (like "Exception ... never thrown"). However, you must either catch Exception itself or add the throws Exception to the calling method to allow passing the exception further upwards. For example,
private void someMethod() throws Exception {
throw new NumberFormatException("Illegal number format");
}
If this method gets called, you can catch the NumberFormatException, but you also have to either handle the more generic Exception or declare it in the throws clause (and then handle it further up in the call hierarchy):
public void myMethod() {
try {
someMethod();
}catch(NumberFormatException nfe) {
nfe.printStackTrace();
}catch(Exception ex) {
ex.printStackTrace();
}
}
Or:
public void myMethod() throws Exception {
try {
someMethod();
}catch(NumberFormatException nfe) {
nfe.printStackTrace();
}
}

Yes, there is a difference.
When you use the throws declaration of an exception in a method, you do it so you can handle it later (using try{}catch blocks).
When you handle the exception, you might just do the e.printStackTrace(), but that's not really handling it.
Instead, imagine you want to tell your user "You didn't introduce a number, please correct this mistake" and prompt them to introduce a number. You can do this if you throw NumberFormatException in the method you use to read. But if you throw just Exception, you can't know for sure if that was the error or any other exception, and you might have unexpected behaviours because of that.

The difference between two approach lies based on Exception Handling. For example, if you like to handle all exception in same way then, you should opt for first approach. On the other hand, if you like to deal different exception in different manner, you must go for second approach

I would prefer to mention the exceptions explicitly in the function declaration after the throws keyword because then I know how I should handle these exceptions if they occur in the calling code. See the code below for better understanding..
public static void exceptionExample() throws SQlException, NamingException {
}
public static void main(String[] args) throws SQlException, NamingException {
try {
exceptionExample();
exceptionExample();
}
catch(SQlException e){
System.out.println("Sql exception has occured");
}
catch(NamingException e){
System.out.println("Naming Exception has occured");
}
}
If I would not have mentioned throws SQlException , NamingException explicitly in the function and have used only throws Exception then it implies I do not know which exception may occur. In that case in the catch() {...} block I would have used e.printStackTrace which is not really an efficient method of exception handling because user may never know what happened to the application. But here as you can see that I can inform the user that the respective exception has occurred which makes my application more user friendly.
Also Throwing Exception causes the calling code to catch the exceptions which they may not want to handle depending upon the application and all kinds of reasons.
So I would say it depends upon the wit of the programmer on how he/she wants to handle the exceptions efficiently.....

what is the major difference between the given 2 programs having TRY & CATCH and THROWS exception in JAVA [duplicate]

So I thought I had a good basic understanding of exception-handling in Java, but I was recently reading some code that gave me some confusion and doubts. My main doubt that I want to address here is when should a person use throws in a Java method declaration like the following:
public void method() throws SomeException
{
// method body here
}
From reading some similar posts I gather that throws is used as a sort of declaration that SomeException could be thrown during the execution of the method.
My confusion comes from some code that looked like this:
public void method() throws IOException
{
try
{
BufferedReader br = new BufferedReader(new FileReader("file.txt"));
}
catch(IOException e)
{
System.out.println(e.getMessage());
}
}
Is there any reason that you would want to use a throws in this example? It seems that if you are just doing basic exception-handling of something like an IOException that you would simply need the try/catch block and that's it.

If you are catching an exception type, you do not need to throw it, unless you are going to rethrow it. In the example you post, the developer should have done one or another, not both.
Typically, if you are not going to do anything with the exception, you should not catch it.
The most dangerous thing you can do is catch an exception and not do anything with it.
A good discussion of when it is appropriate to throw exceptions is here
When to throw an exception?

You only need to include a throws clause on a method if the method throws a checked exception. If the method throws a runtime exception then there is no need to do so.
See here for some background on checked vs unchecked exceptions: http://download.oracle.com/javase/tutorial/essential/exceptions/runtime.html
If the method catches the exception and deals with it internally (as in your second example) then there is no need to include a throws clause.

The code that you looked at is not ideal. You should either:
Catch the exception and handle it;
in which case the throws is
unnecesary.
Remove the try/catch; in which case
the Exception will be handled by a
calling method.
Catch the exception, possibly
perform some action and then rethrow
the exception (not just the message)

You're correct, in that example the throws is superfluous. It's possible that it was left there from some previous implementation - perhaps the exception was originally thrown instead of caught in the catch block.

The code you posted is wrong, it should throw an Exception if is catching a specific exception in order to handler IOException but throwing not catched exceptions.
Something like:
public void method() throws Exception {
try {
BufferedReader br = new BufferedReader(new FileReader("file.txt"));
} catch(IOException e) {
System.out.println(e.getMessage());
}
}
or
public void method() {
try {
BufferedReader br = new BufferedReader(new FileReader("file.txt"));
} catch(IOException e) {
System.out.println("Catching IOException");
System.out.println(e.getMessage());
} catch(Exception e) {
System.out.println("Catching any other Exceptions like NullPontException, FileNotFoundExceptioon, etc.");
System.out.println(e.getMessage());
}
}

In the example you gave, the method will never throw an IOException, therefore the declaration is wrong (but valid). My guess is that the original method threw the IOException, but it was then updated to handle the exception within but the declaration was not changed.

This is not an answer, but a comment, but I could not write a comment with a formatted code, so here is the comment.
Lets say there is
public static void main(String[] args) {
try {
// do nothing or throw a RuntimeException
throw new RuntimeException("test");
} catch (Exception e) {
System.out.println(e.getMessage());
throw e;
}
}
The output is
test
Exception in thread "main" java.lang.RuntimeException: test
at MyClass.main(MyClass.java:10)
That method does not declare any "throws" Exceptions, but throws them!
The trick is that the thrown exceptions are RuntimeExceptions (unchecked) that are not needed to be declared on the method.
It is a bit misleading for the reader of the method, since all she sees is a "throw e;" statement but no declaration of the throws exception
Now, if we have
public static void main(String[] args) throws Exception {
try {
throw new Exception("test");
} catch (Exception e) {
System.out.println(e.getMessage());
throw e;
}
}
We MUST declare the "throws" exceptions in the method otherwise we get a compiler error.

Java: Using Catch and throws in the one block?

What is the point of catching and then also throwing an Exception like below? Is it bad practice to do both?
try{
//something
} catch (Exception e){
throw new RuntimeException("reason for exception");
}

Usually, such code is used to re-wrap exceptions, that means transforming the type of the exception. Typically, you do this when you are limited in what exceptions are allowed out of your method, but internally other types of exceptions can happen. For example:
class MyServiceImplementaiton implements MyService {
void myService() throws MyServiceException { // cannot change the throws clause here
try {
.... // Do something
} catch(IOException e) {
// re-wrap the received IOException as MyServiceException
throw new MyServiceException(e);
}
}
}
This idiom enables to keep propagating exceptions to the caller, while conforming to the throws clause in the interface and hide the details of the internals (the fact that IOExceptions can happen).
In practice, this is always better than just calling e.printStackTrace() which will actually "swallow" the error condition and let the rest of the program run as if nothing had happened. In this respect, behaviour of Eclipse is quite bad as it tends to auto-write such bad-practice constructs if the developer is not careful.

This is called rethrowing an exception, and it is a common pattern.
It allows you to change the class of the exception (such as in this case), or to add more information (also the case here, as long as that error string is meaningful).
It is often a good idea to attach the original exception:
throw new RuntimeException("cause of the problem", e);
Rethrowing as an unchecked exception (a RuntimeException) is sometimes necessary when you still want to throw an exception, but the API of your method does not allow a checked exception.

In your example, an Exception is caught and a RuntimeException is thrown, which effectively replaces a (potentially) checked exception with an unchecked exception that doesn't have to be handled by the caller, nor declared by the throwing method in a throws clause.
Some examples :
This code passes compilation :
public void SomeMethod ()
{
try {
//something
} catch (Exception e){
throw new RuntimeException("reason for exception");
}
}
This code doesn't pass compilation (assuming "something" may throw a checked exception) :
public void SomeMethod ()
{
//something
}
An alternative to catching the Exception and throwing an unchecked exception (i.e. RuntimeException) is to add a throws clause :
public void SomeMethod () throws Exception
{
//something
}
This is one use case of catching one type of exception and throwing another. Another use case is to catch one type of exception and throw another type of checked exception (that your method declares in its throws clause). It is sometimes done in order to group multiple exceptions that may be thrown inside a method, and only throw one type of exception to the caller of the method (which makes it easier for them to write the exception handling code, and makes sense if all those exceptions should be handled in the same manner).