I've tried to throw the same excpetion in a finally block, while the previously throwed expcetion was not catched. I expected that we have two object of Excpetion type that shal be thrown. Since we need in two catch clauses as the following:
public static void main(String[] args) {
try {
try {
try {
throw new Exception();
} finally {
System.out.println("finally");
throw new Exception();
}
} catch (Exception ex) {
System.out.println("catch");
} finally {
System.out.println("finally");
}
} catch (Exception ex) {
System.out.println("catch");
}
System.out.println("finish");
}
But that program prints:
finally
catch
finally
finish
That is, the second catch clause was not entered. Why?
When you throw an exception in the finally block, the first exception silently disappears.
It's in the JLS Chapter 14.20.2
If the finally block completes abruptly for reason S, then the try
statement completes abruptly for reason S.
This is true how ever you entered the finally block. If you entered it by throwing an exception T that exception can not be catched anymore.
When you throw an exception from the finally block, it supresses any exception thrown from the try block, therefore there was only one exception to catch.
The first catch cluase already caught that exception, so the second one had nothing to catch.
Your first try-catch is already trying to catch it, the extra try blocks aren't there for any particular reason. If you try to throw more exceptions you'll notice that you'll get a syntax error, Unreachable code
Basically, keep it in one try block
Related
let's consider the following code:
try {
throw new Exception("from try")
} catch (Exception e) {
throw new Exception("from catch")
} finally {
throw new Exception("from finally")
}
It gives:
Exception thrown
java.lang.Exception: from finally
<...>
So it looks that finally is executed before catch and terminates execution flow.
What could I do if want to see both exceptions?
So it looks that finally is executed before catch and terminates
execution flow.
That is not correct. finally is executed after a correspondeing catch, not before. The issue is that your catch block executes, and then after that the finally block is guaranteed to execute.
This post will help you I guess:
https://stackoverflow.com/questions/3779285/exception-thrown-in-catch-and-finally-clause
My code is below :-
class Varr {
public static void main(String[] args) {
try {
System.out.println(10/0);
}
catch(ArithmeticException e) {
System.out.println("catch1");
System.out.println("catch1");
throw new ArithmeticException ("Exce");
}
finally {
System.out.println("finally");
}
}
}
Output is :-
catch1
catch1
finally
Exception in thread "main" java.lang.ArithmeticException: Exce
at one.Varr.main(Varr.java:22)
As per my knowledge the flow has to be first try then catch and finally at last but as per the output the flow is try then few lines of catch upto the throw exception statement and then finally and the throw exception statement of catch block at last.
Why is there discrepancy in flow, I mean why finally was executed before the throw new exception statement of catch block
Because a block of finally, by definition, has to be executed no matter the outcome of the try or catch clauses.
In your case, the run-time knows there is an exception that has to be propagated upwards, but before doing so, it executes whatever is inside the finally block.
When the finally block is finished, it propagates any exception that might have been raised, or the flow continues otherwise.
You can take a look at the essentials of finally
The finally block always executes when the try block exits. This
ensures that the finally block is executed even if an unexpected
exception occurs.
Here is my code:
public static void main(String[] args) {
System.out.print("a");
try {
System.out.print("b");
throw new IllegalArgumentException();
} catch (IllegalArgumentException e) {
System.out.print("c");
throw new RuntimeException("1");
} catch (RuntimeException e) {
System.out.print("d");
throw new RuntimeException("2");
} finally {
System.out.print("e");
throw new RuntimeException("3");
}
}
I can not understand why the output is abce and RuntimeException("3")
That becomes clear when you indent your code as it should be:
try {
System.out.print("b");
throw new IllegalArgumentException();
} catch (IllegalArgumentException e) {
System.out.print("c");
throw new RuntimeException("1");
} catch (RuntimeException e) {
System.out.print("d");
throw new RuntimeException("2");
} finally {
System.out.print("e");
throw new RuntimeException("3");
}
The point is: there is only one try block. And the first catch block is taken. That one throws - and this exception 1 would be the one you notice in your stacktrace.
But thing is: the finally block is throwing "on top" of exception 1. Therefore you see exception 3.
In other words: there is a misconception on your end. You probably assume that exception 1 should be caught by the second catch block. And that is wrong. The second catch block only covers the first try block. Meaning: an exception from a catch block does not result in another catch block being taken. The first catch block "fires", and the finally block "fires" - leading to the observed results.
While you can have many catch() blocks, a maximum of 1 catch() will be executed when an exception is thrown. After that it will execute finally block, hence you have the output abce and the RuntimeException from finally block.
It is because, after IllegalArgumentException thrown in try block, it is caught in corresponding catch (IllegalArgumentException e) { } block. As finally block gets executed regardless of exception, e get printed.
To your question, Since the exception was already caught in catch block it will throw the corresponding exception accordingly and it will get propagated to the caller of that method.
class chain_exceptions{
public static void main(String args[]){
try
{
f1();
}
catch(IndexOutOfBoundsException e)
{
System.out.println("A");
throw new NullPointerException(); //Line 0
}
catch(NullPointerException e) //Line 1
{
System.out.println("B");
return;
}
catch (Exception e)
{
System.out.println("C");
}
finally
{
System.out.println("D");
}
System.out.println("E");
}
static void f1(){
System.out.println("Start...");
throw new IndexOutOfBoundsException( "parameter" );
}
}
I expected the Line 1 to catch the NullPointerException thrown from the Line 0 but it does not happen.
But why so ?.
When there is another catch block defined, why cant the NPE handler at Line1 catch it ?
Is it because the "throw" goes directly to the main() method ?
Catch{ . . . } blocks are associated with try{ . . . } blocks. A catch block can catch exceptions that are thrown from a try block only. The other catch blocks after first catch block are not associated with a try block and hence when you throw an exception, they are not caught. Or main() does not catch exceptions.
A kind of this for each catch block will do what you are trying to do.
try{
try
{
f1();
}
catch(IndexOutOfBoundsException e)
{
System.out.println("A");
throw new NullPointerException(); //Line 0
}
}
catch(NullPointerException e) //Line 1
{
System.out.println("B");
return;
}
The catch blocks are only for the try block. They won't catch exceptions from other catch blocks.
catch statements only catch exceptions thrown from a try { ... } block.
The NullPointerException is thrown from a catch { ... } block, not a try { ... } block.
To catch an exception thrown from a catch block you need to put another try block inside of it. Outside, wrapping the original try...catch would work, too.
A second catch doesn't catch the exception from the first catch block. You have to add another try-catch within the first catch block (or around the whole try-catch you already have) to make this run as expected.
Since java 7 you can use code below or an other option is to nest the try catch statements, there is no other option in java
try {
...
} catch( indexoutofboundsexception| nullpointerexception ex ) {
logger.log(ex);
throw ex;
}
Your catch clauses only catch exceptions thrown by f1(). They don't call exceptions thrown in other catch clauses of the same try-catch-finally construct.
Because f1() throws IndexOutOfBoundsException.
try
{
f1(); //throws IndexOutOfBoundsException
}
catch(IndexOutOfBoundsException e) //gets caught here immediately and does not check other catch blocks
{
System.out.println("A");
throw new NullPointerException(); //Line 0
}
Short answer: yes, the throw will directly throw the exception to the main method.
Generally, once a catch block is executed, it behaves like an else if, that is, it won't consider the other alternatives.
No, the reason it isn't being caught is because it isn't in the try block which is linked to the catch block. If you want to catch that exception as well, you would have to wrap the throw in a new try/catch group. The reason why you would want to do this tho, is a riddle to me.
What you also can do btw:
catch (IndexOutOfBoundsException|NullPointerException e)
This will also allow you to use the same catch block for multiple types of exceptions.
Your expectation was incorrect:
The catch blocks are associated with the try block. So once an exception is thrown inside of the try, it leaves that scope. Now you are in the scope outside the try, meaning you are no longer in any try/catch block. Any exceptions thrown here (when you re-throw) will not be caught by anything, and yes, bubble out of main.
You can not catch exception from another catch block, for that you probably need to do something like this, in your first catch block
System.out.println("A");
try{
throw new NullPointerException(); //Line 0
}
catch(NullPointerException e) //Line 1
{
System.out.println("B");
return;
}
On a question for Java at the university, there was this snippet of code:
class MyExc1 extends Exception {}
class MyExc2 extends Exception {}
class MyExc3 extends MyExc2 {}
public class C1 {
public static void main(String[] args) throws Exception {
try {
System.out.print(1);
q();
}
catch (Exception i) {
throw new MyExc2();
}
finally {
System.out.print(2);
throw new MyExc1();
}
}
static void q() throws Exception {
try {
throw new MyExc1();
}
catch (Exception y) {
}
finally {
System.out.print(3);
throw new Exception();
}
}
}
I was asked to give its output. I answered 13Exception in thread main MyExc2, but the correct answer is 132Exception in thread main MyExc1. Why is it that? I just can't understand where does MyExc2 go.
Based on reading your answer and seeing how you likely came up with it, I believe you think an "exception-in-progress" has "precedence". Keep in mind:
When an new exception is thrown in a catch block or finally block that will propagate out of that block, then the current exception will be aborted (and forgotten) as the new exception is propagated outward. The new exception starts unwinding up the stack just like any other exception, aborting out of the current block (the catch or finally block) and subject to any applicable catch or finally blocks along the way.
Note that applicable catch or finally blocks includes:
When a new exception is thrown in a catch block, the new exception is still subject to that catch's finally block, if any.
Now retrace the execution remembering that, whenever you hit throw, you should abort tracing the current exception and start tracing the new exception.
Exceptions in the finally block supersede exceptions in the catch block.
Java Language Specification
If the catch block completes abruptly for reason R, then the finally
block is executed. Then there is a choice:
If the finally block completes normally, then the try statement completes abruptly for reason R.
If the finally block completes abruptly for reason S, then the try statement completes abruptly for reason S (and reason R is discarded).
This is what Wikipedia says about finally clause:
More common is a related clause
(finally, or ensure) that is executed
whether an exception occurred or not,
typically to release resources
acquired within the body of the
exception-handling block.
Let's dissect your program.
try {
System.out.print(1);
q();
}
So, 1 will be output into the screen, then q() is called. In q(), an exception is thrown. The exception is then caught by Exception y but it does nothing. A finally clause is then executed (it has to), so, 3 will be printed to screen. Because (in method q() there's an exception thrown in the finally clause, also q() method passes the exception to the parent stack (by the throws Exception in the method declaration) new Exception() will be thrown and caught by catch ( Exception i ), MyExc2 exception will be thrown (for now add it to the exception stack), but a finally in the main block will be executed first.
So in,
catch ( Exception i ) {
throw( new MyExc2() );
}
finally {
System.out.print(2);
throw( new MyExc1() );
}
A finally clause is called...(remember, we've just caught Exception i and thrown MyExc2) in essence, 2 is printed on screen...and after the 2 is printed on screen, a MyExc1 exception is thrown. MyExc1 is handled by the public static void main(...) method.
Output:
"132Exception in thread main MyExc1"
Lecturer is correct! :-)
In essence, if you have a finally in a try/catch clause, a finally will be executed (after catching the exception before throwing the caught exception out)
Finally clause is executed even when exception is thrown from anywhere in try/catch block.
Because it's the last to be executed in the main and it throws an exception, that's the exception that the callers see.
Hence the importance of making sure that the finally clause does not throw anything, because it can swallow exceptions from the try block.
A method can't throw two exceptions at the same time. It will always throw the last thrown exception, which in this case it will be always the one from the finally block.
When the first exception from method q() is thrown, it will catch'ed and then swallowed by the finally block thrown exception.
q() -> thrown new Exception -> main catch Exception -> throw new Exception -> finally throw a new exception (and the one from the catch is "lost")
class MyExc1 extends Exception {}
class MyExc2 extends Exception {}
class MyExc3 extends MyExc2 {}
public class C1 {
public static void main(String[] args) throws Exception {
try {
System.out.print("TryA L1\n");
q();
System.out.print("TryB L1\n");
}
catch (Exception i) {
System.out.print("Catch L1\n");
}
finally {
System.out.print("Finally L1\n");
throw new MyExc1();
}
}
static void q() throws Exception {
try {
System.out.print("TryA L2\n");
q2();
System.out.print("TryB L2\n");
}
catch (Exception y) {
System.out.print("Catch L2\n");
throw new MyExc2();
}
finally {
System.out.print("Finally L2\n");
throw new Exception();
}
}
static void q2() throws Exception {
throw new MyExc1();
}
}
Order:
TryA L1
TryA L2
Catch L2
Finally L2
Catch L1
Finally L1
Exception in thread "main" MyExc1 at C1.main(C1.java:30)
https://www.compilejava.net/
The logic is clear till finish printing out 13. Then the exception thrown in q() is caught by catch (Exception i) in main() and a new MyEx2() is ready to be thrown. However, before throwing the exception, the finally block have to be executed first. Then the output becomes 132 and finally asks to thrown another exception new MyEx1().
As a method cannot throw more than one Exception, it will always throw the latest Exception. In other words, if both catch and finally blocks try to throw Exception, then the Exception in catch is swallowed and only the exception in finally will be thrown.
Thus, in this program, Exception MyEx2 is swallowed and MyEx1 is thrown. This Exception is thrown out of main() and no longer caught, thus JVM stops and the final output is 132Exception in thread main MyExc1.
In essence, if you have a finally in a try/catch clause, a finally will be executed AFTER catching the exception, but BEFORE throwing any caught exception, and ONLY the lastest exception would be thrown in the end.
The easiest way to think of this is imagine that there is a variable global to the entire application that is holding the current exception.
Exception currentException = null;
As each exception is thrown, "currentException" is set to that exception. When the application ends, if currentException is != null, then the runtime reports the error.
Also, the finally blocks always run before the method exits. You could then requite the code snippet to:
public class C1 {
public static void main(String [] argv) throws Exception {
try {
System.out.print(1);
q();
}
catch ( Exception i ) {
// <-- currentException = Exception, as thrown by q()'s finally block
throw( new MyExc2() ); // <-- currentException = MyExc2
}
finally {
// <-- currentException = MyExc2, thrown from main()'s catch block
System.out.print(2);
throw( new MyExc1() ); // <-- currentException = MyExc1
}
} // <-- At application exit, currentException = MyExc1, from main()'s finally block. Java now dumps that to the console.
static void q() throws Exception {
try {
throw( new MyExc1() ); // <-- currentException = MyExc1
}
catch( Exception y ) {
// <-- currentException = null, because the exception is caught and not rethrown
}
finally {
System.out.print(3);
throw( new Exception() ); // <-- currentException = Exception
}
}
}
The order in which the application executes is:
main()
{
try
q()
{
try
catch
finally
}
catch
finally
}
It is well known that the finally block is executed after the the try and catch and is always executed....
But as you saw it's a little bit tricky sometimes check out those code snippet below and you will that
the return and throw statements don't always do what they should do in the order that we expect theme to.
Cheers.
/////////////Return dont always return///////
try{
return "In Try";
}
finally{
return "In Finally";
}
////////////////////////////////////////////
////////////////////////////////////////////
while(true) {
try {
return "In try";
}
finally{
break;
}
}
return "Out of try";
///////////////////////////////////////////
///////////////////////////////////////////////////
while (true) {
try {
return "In try";
}
finally {
continue;
}
}
//////////////////////////////////////////////////
/////////////////Throw dont always throw/////////
try {
throw new RuntimeException();
}
finally {
return "Ouuuups no throw!";
}
//////////////////////////////////////////////////
I think you just have to walk the finally blocks:
Print "1".
finally in q print "3".
finally in main print "2".
I think this solve the problem :
boolean allOk = false;
try{
q();
allOk = true;
} finally {
try {
is.close();
} catch (Exception e) {
if(allOk) {
throw new SomeException(e);
}
}
}
To handle this kind of situation i.e. handling the exception raised by finally block. You can surround the finally block by try block: Look at the below example in python:
try:
fh = open("testfile", "w")
try:
fh.write("This is my test file for exception handling!!")
finally:
print "Going to close the file"
fh.close()
except IOError:
print "Error: can\'t find file or read data"