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
I am trying to put a while loop inside a try /catch block. To my curiosity finally of that try catch is not executed when the while loop exits. Can some explain what is happening actually?
I tried to google, but cnould not find any details.
I assume your code looks like this:
try
{
while (...)
{
// ...
}
}
catch (FooException ex)
{
// This only executes if a FooException is thrown.
}
finally
{
// This executes whether or not there is an exception.
}
The catch block is only executed if there is an exception. The finally block usually executes whether or not an exception was thrown. So you will probably find that your finally block is actually being executed. You can prove this to yourself by placing a line there that causes some output to the console.
However there are situations in which the finally block doesn't run. See here for more details:
Does a finally block always run?
It can happen only if your program exits either by using System.exit() or if an Error or Throwable were thrown (vs. Exception that will be caught).
Try the following:
public static void main(String[] args) {
try{
System.out.println("START!");
int i=0;
while(true){
i++;
if(i > 10){
System.exit(1);
}
}
}
catch (Exception e) {
// TODO: handle exception
}
finally{
System.out.println("this will not be printed!");
}
}
Considering the below code, does finally still run if there was an exception thrown in the catch clause?
try {
//code here throws exception
}
catch(Exception ex) {
//code catches above exception however code here also throws another exception
}
finally {
//does this code even run considering the exception thrown in the above catch clause??
}
Yes it does. It runs regardless of what happens in the try/catch (assuming the JVM doesn't shut down for some reason)
In readFileMethod1, an IOException is explicitly catched before throwing it at the method level to ensure that the finally block is executed. However, is it neccessary to catch the exception? If I remove the catch block, shown in readFileMethod2, does the finally block get executed as well?
private void readFileMethod1() throws IOException {
try {
// do some IO stuff
} catch (IOException ex) {
throw ex;
} finally {
// release resources
}
}
private void readFileMethod2() throws IOException {
try {
// do some IO stuff
} finally {
// release resources
}
}
The finally still gets executed, regardless of whether you catch the IOException. If all your catch block does is rethrow, then it is not necessary here.
No, it's completely unnecessary to catch an exception if you're not going to do anything other than throw it.
And yes, the finally block will still be executed.
No, it isn't necessary to catch the exception unless you can't rethrow it in your method. In the code you posted the readFileMethod2 is the correct option to follow.
the finally always get executed in try catch context ... for more info check http://download.oracle.com/javase/tutorial/essential/exceptions/finally.html
finally is executed always irrespective of whether an exception is thrown or not. Only if the JVM is shut down while executing the try block or catch block, then the finally clause will not be executed. Likewise, if the thread executing the try or catch code is interrupted or killed, the finally block may not execute even though the application as a whole continues.
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"