Considering this code, can I be absolutely sure that the finally block always executes, no matter what something() is?
try {
something();
return success;
}
catch (Exception e) {
return failure;
}
finally {
System.out.println("I don't know if this will get printed out");
}
Yes, finally will be called after the execution of the try or catch code blocks.
The only times finally won't be called are:
If you invoke System.exit()
If you invoke Runtime.getRuntime().halt(exitStatus)
If the JVM crashes first
If the JVM reaches an infinite loop (or some other non-interruptable, non-terminating statement) in the try or catch block
If the OS forcibly terminates the JVM process; e.g., kill -9 <pid> on UNIX
If the host system dies; e.g., power failure, hardware error, OS panic, et cetera
If the finally block is going to be executed by a daemon thread and all other non-daemon threads exit before finally is called
Example code:
public static void main(String[] args) {
System.out.println(Test.test());
}
public static int test() {
try {
return 0;
}
finally {
System.out.println("something is printed");
}
}
Output:
something is printed.
0
Also, although it's bad practice, if there is a return statement within the finally block, it will trump any other return from the regular block. That is, the following block would return false:
try { return true; } finally { return false; }
Same thing with throwing exceptions from the finally block.
Here's the official words from the Java Language Specification.
14.20.2. Execution of try-finally and try-catch-finally
A try statement with a finally block is executed by first executing the try block. Then there is a choice:
If execution of the try block completes normally, [...]
If execution of the try block completes abruptly because of a throw of a value V, [...]
If execution of the try block completes abruptly for any other 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).
The specification for return actually makes this explicit:
JLS 14.17 The return Statement
ReturnStatement:
return Expression(opt) ;
A return statement with no Expression attempts to transfer control to the invoker of the method or constructor that contains it.
A return statement with an Expression attempts to transfer control to the invoker of the method that contains it; the value of the Expression becomes the value of the method invocation.
The preceding descriptions say "attempts to transfer control" rather than just "transfers control" because if there are any try statements within the method or constructor whose try blocks contain the return statement, then any finally clauses of those try statements will be executed, in order, innermost to outermost, before control is transferred to the invoker of the method or constructor. Abrupt completion of a finally clause can disrupt the transfer of control initiated by a return statement.
In addition to the other responses, it is important to point out that 'finally' has the right to override any exception/returned value by the try..catch block. For example, the following code returns 12:
public static int getMonthsInYear() {
try {
return 10;
}
finally {
return 12;
}
}
Similarly, the following method does not throw an exception:
public static int getMonthsInYear() {
try {
throw new RuntimeException();
}
finally {
return 12;
}
}
While the following method does throw it:
public static int getMonthsInYear() {
try {
return 12;
}
finally {
throw new RuntimeException();
}
}
Here's an elaboration of Kevin's answer. It's important to know that the expression to be returned is evaluated before finally, even if it is returned after.
public static void main(String[] args) {
System.out.println(Test.test());
}
public static int printX() {
System.out.println("X");
return 0;
}
public static int test() {
try {
return printX();
}
finally {
System.out.println("finally trumps return... sort of");
return 42;
}
}
Output:
X
finally trumps return... sort of
42
I tried the above example with slight modification-
public static void main(final String[] args) {
System.out.println(test());
}
public static int test() {
int i = 0;
try {
i = 2;
return i;
} finally {
i = 12;
System.out.println("finally trumps return.");
}
}
The above code outputs:
finally trumps return.
2
This is because when return i; is executed i has a value 2. After this the finally block is executed where 12 is assigned to i and then System.out out is executed.
After executing the finally block the try block returns 2, rather than returning 12, because this return statement is not executed again.
If you will debug this code in Eclipse then you'll get a feeling that after executing System.out of finally block the return statement of try block is executed again. But this is not the case. It simply returns the value 2.
That is the whole idea of a finally block. It lets you make sure you do cleanups that might otherwise be skipped because you return, among other things, of course.
Finally gets called regardless of what happens in the try block (unless you call System.exit(int) or the Java Virtual Machine kicks out for some other reason).
A logical way to think about this is:
Code placed in a finally block must be executed whatever occurs within the try block
So if code in the try block tries to return a value or throw an exception the item is placed 'on the shelf' till the finally block can execute
Because code in the finally block has (by definition) a high priority it can return or throw whatever it likes. In which case anything left 'on the shelf' is discarded.
The only exception to this is if the VM shuts down completely during the try block e.g. by 'System.exit'
finally is always executed unless there is abnormal program termination (like calling System.exit(0)..). so, your sysout will get printed
No, not always one exception case is//
System.exit(0);
before the finally block prevents finally to be executed.
class A {
public static void main(String args[]){
DataInputStream cin = new DataInputStream(System.in);
try{
int i=Integer.parseInt(cin.readLine());
}catch(ArithmeticException e){
}catch(Exception e){
System.exit(0);//Program terminates before executing finally block
}finally{
System.out.println("Won't be executed");
System.out.println("No error");
}
}
}
Also a return in finally will throw away any exception. http://jamesjava.blogspot.com/2006/03/dont-return-in-finally-clause.html
The finally block is always executed unless there is abnormal program termination, either resulting from a JVM crash or from a call to System.exit(0).
On top of that, any value returned from within the finally block will override the value returned prior to execution of the finally block, so be careful of checking all exit points when using try finally.
Finally is always run that's the whole point, just because it appears in the code after the return doesn't mean that that's how it's implemented. The Java runtime has the responsibility to run this code when exiting the try block.
For example if you have the following:
int foo() {
try {
return 42;
}
finally {
System.out.println("done");
}
}
The runtime will generate something like this:
int foo() {
int ret = 42;
System.out.println("done");
return 42;
}
If an uncaught exception is thrown the finally block will run and the exception will continue propagating.
NOT ALWAYS
The Java Language specification describes how try-catch-finally and try-catch blocks work at 14.20.2
In no place it specifies that the finally block is always executed.
But for all cases in which the try-catch-finally and try-finally blocks complete it does specify that before completion finally must be executed.
try {
CODE inside the try block
}
finally {
FIN code inside finally block
}
NEXT code executed after the try-finally block (may be in a different method).
The JLS does not guarantee that FIN is executed after CODE.
The JLS guarantees that if CODE and NEXT are executed then FIN will always be executed after CODE and before NEXT.
Why doesn't the JLS guarantee that the finally block is always executed after the try block? Because it is impossible. It is unlikely but possible that the JVM will be aborted (kill, crash, power off) just after completing the try block but before execution of the finally block. There is nothing the JLS can do to avoid this.
Thus, any software which for their proper behaviour depends on finally blocks always being executed after their try blocks complete are bugged.
return instructions in the try block are irrelevant to this issue. If execution reaches code after the try-catch-finally it is guaranteed that the finally block will have been executed before, with or without return instructions inside the try block.
Yes it will get called. That's the whole point of having a finally keyword. If jumping out of the try/catch block could just skip the finally block it was the same as putting the System.out.println outside the try/catch.
Because a finally block will always be called unless you call System.exit() (or the thread crashes).
Concisely, in the official Java Documentation (Click here), it is written that -
If the JVM exits while the try or catch code is being executed, then
the finally block may not execute. 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.
This is because you assigned the value of i as 12, but did not return the value of i to the function. The correct code is as follows:
public static int test() {
int i = 0;
try {
return i;
} finally {
i = 12;
System.out.println("finally trumps return.");
return i;
}
}
Answer is simple YES.
INPUT:
try{
int divideByZeroException = 5 / 0;
} catch (Exception e){
System.out.println("catch");
return; // also tried with break; in switch-case, got same output
} finally {
System.out.println("finally");
}
OUTPUT:
catch
finally
finally block is always executed and before returning x's (calculated) value.
System.out.println("x value from foo() = " + foo());
...
int foo() {
int x = 2;
try {
return x++;
} finally {
System.out.println("x value in finally = " + x);
}
}
Output:
x value in finally = 3
x value from foo() = 2
Yes, it will. No matter what happens in your try or catch block unless otherwise System.exit() called or JVM crashed. if there is any return statement in the block(s),finally will be executed prior to that return statement.
Adding to #vibhash's answer as no other answer explains what happens in the case of a mutable object like the one below.
public static void main(String[] args) {
System.out.println(test().toString());
}
public static StringBuffer test() {
StringBuffer s = new StringBuffer();
try {
s.append("sb");
return s;
} finally {
s.append("updated ");
}
}
Will output
sbupdated
Yes It will.
Only case it will not is JVM exits or crashes
Yes, finally block is always execute. Most of developer use this block the closing the database connection, resultset object, statement object and also uses into the java hibernate to rollback the transaction.
finally will execute and that is for sure.
finally will not execute in below cases:
case 1 :
When you are executing System.exit().
case 2 :
When your JVM / Thread crashes.
case 3 :
When your execution is stopped in between manually.
I tried this,
It is single threaded.
public static void main(String args[]) throws Exception {
Object obj = new Object();
try {
synchronized (obj) {
obj.wait();
System.out.println("after wait()");
}
} catch (Exception ignored) {
} finally {
System.out.println("finally");
}
}
The main Thread will be on wait state forever, hence finally will never be called,
so console output will not print String: after wait() or finally
Agreed with #Stephen C, the above example is one of the 3rd case mention here:
Adding some more such infinite loop possibilities in following code:
// import java.util.concurrent.Semaphore;
public static void main(String[] args) {
try {
// Thread.sleep(Long.MAX_VALUE);
// Thread.currentThread().join();
// new Semaphore(0).acquire();
// while (true){}
System.out.println("after sleep join semaphore exit infinite while loop");
} catch (Exception ignored) {
} finally {
System.out.println("finally");
}
}
Case 2: If the JVM crashes first
import sun.misc.Unsafe;
import java.lang.reflect.Field;
public static void main(String args[]) {
try {
unsafeMethod();
//Runtime.getRuntime().halt(123);
System.out.println("After Jvm Crash!");
} catch (Exception e) {
} finally {
System.out.println("finally");
}
}
private static void unsafeMethod() throws NoSuchFieldException, IllegalAccessException {
Field f = Unsafe.class.getDeclaredField("theUnsafe");
f.setAccessible(true);
Unsafe unsafe = (Unsafe) f.get(null);
unsafe.putAddress(0, 0);
}
Ref: How do you crash a JVM?
Case 6: If finally block is going to be executed by daemon Thread and all other non-daemon Threads exit before finally is called.
public static void main(String args[]) {
Runnable runnable = new Runnable() {
#Override
public void run() {
try {
printThreads("Daemon Thread printing");
// just to ensure this thread will live longer than main thread
Thread.sleep(10000);
} catch (Exception e) {
} finally {
System.out.println("finally");
}
}
};
Thread daemonThread = new Thread(runnable);
daemonThread.setDaemon(Boolean.TRUE);
daemonThread.setName("My Daemon Thread");
daemonThread.start();
printThreads("main Thread Printing");
}
private static synchronized void printThreads(String str) {
System.out.println(str);
int threadCount = 0;
Set<Thread> threadSet = Thread.getAllStackTraces().keySet();
for (Thread t : threadSet) {
if (t.getThreadGroup() == Thread.currentThread().getThreadGroup()) {
System.out.println("Thread :" + t + ":" + "state:" + t.getState());
++threadCount;
}
}
System.out.println("Thread count started by Main thread:" + threadCount);
System.out.println("-------------------------------------------------");
}
output: This does not print "finally" which implies "Finally block" in "daemon thread" did not execute
main Thread Printing
Thread :Thread[My Daemon Thread,5,main]:state:BLOCKED
Thread :Thread[main,5,main]:state:RUNNABLE
Thread :Thread[Monitor Ctrl-Break,5,main]:state:RUNNABLE
Thread count started by Main thread:3
-------------------------------------------------
Daemon Thread printing
Thread :Thread[My Daemon Thread,5,main]:state:RUNNABLE
Thread :Thread[Monitor Ctrl-Break,5,main]:state:RUNNABLE
Thread count started by Main thread:2
-------------------------------------------------
Process finished with exit code 0
Consider the following program:
public class SomeTest {
private static StringBuilder sb = new StringBuilder();
public static void main(String args[]) {
System.out.println(someString());
System.out.println("---AGAIN---");
System.out.println(someString());
System.out.println("---PRINT THE RESULT---");
System.out.println(sb.toString());
}
private static String someString() {
try {
sb.append("-abc-");
return sb.toString();
} finally {
sb.append("xyz");
}
}
}
As of Java 1.8.162, the above code block gives the following output:
-abc-
---AGAIN---
-abc-xyz-abc-
---PRINT THE RESULT---
-abc-xyz-abc-xyz
this means that using finally to free up objects is a good practice like the following code:
private static String someString() {
StringBuilder sb = new StringBuilder();
try {
sb.append("abc");
return sb.toString();
} finally {
sb = null; // Just an example, but you can close streams or DB connections this way.
}
}
That's actually true in any language...finally will always execute before a return statement, no matter where that return is in the method body. If that wasn't the case, the finally block wouldn't have much meaning.
In addition to the point about return in finally replacing a return in the try block, the same is true of an exception. A finally block that throws an exception will replace a return or exception thrown from within the try block.
Here's an example from the book "Java All-in-one desk reference"
public class CrazyWithZeros {
public static void main(String[] args) {
try {
int answer = divideTheseNumbers(5, 0);
} catch (Exception e) {
System.out.println("Tried twice, still didn't work!");
}
}
public static int divideTheseNumbers(int a, int b) throws Exception {
int c;
try {
c = a / b;
System.out.println("It worked!");
} catch (Exception e) {
System.out.println("Didn't work the first time.");
c = a / b;
System.out.println("It worked the second time!");
} finally {
System.out.println("Better clean up my mess.");
}
System.out.println("It worked after all.");
return c;
}
}
After the finally clause executes, the ArithmeticException is thrown back up to the calling method. The statement System.out.println("It worked after all.");would never be executed in this case. But what happened to the return c;?
I wonder whether the return statement would still return the result of the division or not?
========
I tried to replace "System.out.println("Better clean up my mess.");" with "System.out.println(c);", then it's compiled and the results are as follows:
Didn't work the first time.
0
Tried twice, still didn't work!
I can't believe the variable c could be calculated. (it's the wrong number, though) Why could this happen?
Then I also tried to replace "System.out.println("Better clean up my mess.");" with "return c;" and deleted the statements below the finally block, it's compiled again...Since the finally block is executed whether or not any exceptions are thrown by the try block or caught by any catch blocks, the return c; should be executed. But here're the results:
Didn't work the first time.
looks like c couldn't get returned...
return c is not executed either. It goes straight to the catch block in your main method.
What do you expect performing an error-prone operation the second time? :)
It is gonna generate an exception of the same type you came in the catch block with, but at that time it would not be handled - you don't have another try-catch within this catch block.
The finally is executed always regardless either an exception occurs or a normal process flow proceeds. In your case, you come to the finally block with the exception and throw it to the caller (main) where it gets handled by its own catch block.
I wonder whether the return statement would still return the result of the division or not?
What do you want to return? You haven't initialized the variable c and there is no correct record to this variable. Therefore, Java doesn't allow to write "something unexpected or unpredictable" into the c.
A method returns some value if it is executed without exception or error.
finally block does not have any impact on whether return statement will be executed or not. (of course, finally block should not throw any further exception)
catch block determines whether that exception should be propagated further or handled within the method.
In the given example, an ArithmeticException is thrown from try block and it will be handled by respective catch block. As catch blocks again throws the same exception, given return statement would never execute.
In short, return c; is never executed in above program and variable c will be deleted as any other local variable.
I was going through the following code:
public int returnSomething() {
try {
throw new RuntimeException("foo!");
} finally {
return 0;
}
}
Please explain what this piece of code is doing. My analysis is that we are throwing a runtime exception inside the method, but after that, the "finally" block will definitely execute. Is that correct?
The answer is in the Java Language Specification section 14.20.2. You need to be aware that returning counts as "completing abruptly":
...
If execution of the try block completes abruptly because of a throw of a value V, then there is a choice
[...]
If the finally block completes abruptly for any reason, then the try statement completes abruptly for the same reason.
(All paths have that same final point, with some slightly different wording.)
So the overall result is that 0 is returned and the exception is discarded.
It's rarely a good idea to return from a finally block.
The contents of the finally block always gets executed. The only few reasons I know it won't execute is when you pull the plug or your JVM crashes. SO I reckon this will return 0.
A quick test shows that this returns 0, instead of propagating the RuntimeException.
public class Test {
public static int returnSomething() {
try {
throw new RuntimeException("foo!");
} finally {
return 0;
}
}
public static void main(String[] args) {
int i=returnSomething();
System.out.println(i);
}
}
I run this code:
public class User {
public static void main(String args[]) {
int array[] = new int[10];
int i = 1;
try {
System.out.println("try: " + i++);
System.out.println(array[10]);
System.out.println("try");
} catch (Exception e) {
System.out.println("catch: " + i++);
System.out.println(array[10]);
System.out.println("catch");
} finally {
System.out.println("finally: " + i++);
Object o = null;
o.hashCode();
System.out.println("finally");
}
}
}
Result: try: 1 catch: 2
finally: 3 Exception in thread
"main" java.lang.NullPointerException
at user.main(User.java:17)
in block catch - ArrayIndexOutOfBoundsException, but we loss this Exception, why?
From the JLS
You can read about this in the JLS, Blocks and Statements, section "14.19.2 Execution of try-catch-finally". And I quote,
If execution of the try block completes abruptly for any other 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). The example...
Therefore, the following (which is really condensed from the questioner's code) completes with an NPE, not the ExceptionTest thrown.
class Phinally
{
static class ExceptionTest extends Exception
{ public ExceptionTest(String message) { super(message); } }
public static void main(String[] args) throws ExceptionTest
{
try {
System.out.println("Foo.");
throw new ExceptionTest("throw from try");
} finally {
throw new NullPointerException("throw from finally");
}
}
}
A Sidebar about try with resources/ARM blocks
Difficulties reasoning about this in some common cases specifically with managing resources, and requiring nested try/catch/finally blocks, and nested inside finally blocks, are part of the reason for the "try with resource" feature in project COIN (to be integrated into Java "fairly soon"), about which you can read more about here.
This is one of many good reasons to invest the time in running a static analyzer like PMD, which finds and complains about this type of confusion -- though it might not catch the case in your code, I'm not sure.
Static Checking
Follow up on comment from #stacktrace: I ran the relevant code through PMD and FindBugs, trying both of the following:
finally { throw NullPointerException("Foo"); }
and
finally { Object o = null; System.out.println(o.toString()); }
For the former, PMD noticed and complained about an exception being thrown from a finally clause. FindBugs doesn't complain at all. For the latter, PMD complained about several things but nothing related ("LocalVariableCouldBeFinal", "StringToString", and "UselessOperationOnImmutable"). However, FindBugs noticed and complained about a null dereference. Moral of the story? Run both PMD and FindBugs!
Related
Related on SO: Swallowing exception thrown in catch/finally. Can I avoid such cumbersome try/catch/finally...
You have just stumbled upon an odd feature of Java, that if the finally block doesn't terminate properly, it hides any exceptions that were previously thrown.
This is by design, it is not a mistake.
You last exception doesn't occur inside a try { } catch { } block so there is no catch { } or finally { } processing.
This question already has answers here:
Why use finally
(9 answers)
Closed 1 year ago.
As far as I can tell, both of the following code snippets will serve the same purpose. Why have finally blocks at all?
Code A:
try { /* Some code */ }
catch { /* Exception handling code */ }
finally { /* Cleanup code */ }
Code B:
try { /* Some code */ }
catch { /* Exception handling code */ }
// Cleanup code
What happens if an exception you're not handling gets thrown? (I hope you're not catching Throwable...)
What happens if you return from inside the try block?
What happens if the catch block throws an exception?
A finally block makes sure that however you exit that block (modulo a few ways of aborting the whole process explicitly), it will get executed. That's important for deterministic cleanup of resources.
Note that (in Java at least, probably also in C#) it's also possible to have a try block without a catch, but with a finally. When an exception happens in the try block, the code in the finally block is run before the exception is thrown higher up:
InputStream in = new FileInputStream("somefile.xyz");
try {
somethingThatMightThrowAnException();
}
finally {
// cleanup here
in.close();
}
You may want to put the code that you want to anyway get executed irrespective of what happens in your try or catch block.
Also if you are using multiple catch and if you want to put some code which is common for all the catch blocks this would be a place to put- but you cannot be sure that the entire code in try has been executed.
For example:
conn c1 = new connection();
try {
c1.dosomething();
} catch (ExceptionA exa) {
handleexA();
//c1.close();
} catch (ExceptionB exb) {
handleexB();
//c1.close();
} finally {
c1.close();
}
Finally always gets executed, where as your code after the catch may not.
Even though our application is closed forcefully there will be some tasks, which we must execute (like memory release, closing database, release lock, etc), if you write these lines of code in the finally block it will execute whether an exception is thrown or not...
Your application may be a collection of threads, Exception terminates the thread but not the whole application, in this case finally is more useful.
In some cases finally won't execute such as JVM Fail, Thread terminate, etc.
Still scrolling down? Here you go!
This question gave me tough time back a while.
try
{
int a=1;
int b=0;
int c=a/b;
}
catch(Exception ex)
{
console.writeline(ex.Message);
}
finally
{
console.writeline("Finally block");
}
console.writeline("After finally");
what would be printed in the above scenario?
Yes guessed it right:
ex.Message--whatever it is (probably attempted division by zero)
Finally block
After finally
try
{
int a=1;
int b=0;
int c=a/b;
}
catch(Exception ex)
{
throw(ex);
}
finally
{
console.writeline("Finally block");
}
console.writeline("After finally");
What would this print? Nothing! It throws an error since the catch block raised an error.
In a good programming structure, your exceptions would be funneled, in the sense that this code will be handled from another layer. To stimulate such a case i'll nested try this code.
try
{
try
{
int a=1;
int b=0;
int c=a/b;
}
catch(Exception ex)
{
throw(ex);
}
finally
{
console.writeline("Finally block")
}
console.writeline("After finally");
}
catch(Exception ex)
{
console.writeline(ex.Message);
}
In this case the output would be:
Finally block
ex.Message--whatever it is.
It is clear that when you catch an exception and throw it again into other layers(Funneling), the code after throw does not get executed. It acts similar to just how a return inside a function works.
You now know why not to close your resources on codes after the catch block.Place them in finally block.
Because you need that code to execute regardless of any exceptions that may be thrown. For example, you may need to clean up some unmanaged resource (the 'using' construct compiles to a try/finally block).
There may be times when you want to execute a piece of code no matter what. Whether an exception is thrown or not. Then one uses finally.
finally ALWAYS executes, unless the JVM was shut down, finally just provides a method to put the cleanup code in one place.
It would be too tedious if you had to put the clean up code in each of the catch blocks.
If catch block throws any exception then remaining code will not executed hence we have to write finaly block.
finally block in java can be used to put "cleanup" code such as closing a file, closing connection etc.
The finally block will not be executed if program exits(either by calling System.exit() or by causing a fatal error that causes the process to abort).