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try and Finally giving exception with no return statement , but there is no exception when return statement is written in method
(3 answers)
Closed 8 years ago.
Trying out a try catch finally use case -
public class Finallyy {
public static void main(String[] args) {
int result = method1();
System.out.println(result);
}
private static int method1() {
int a = 2;
try {
int b = 0;
a = a / a;
System.out.println("try");
return a;
} catch (Exception e) {
System.out.println("caught");
a = 5;
return a;
}
finally {
System.out.println("finally");
a = 10;
return a;
}
}
}
Output:
try
finally
10
If the code is modified to
finally {
System.out.println("finally");
a = 10;
//return a;
}
Output:
try
finally
1
Question - Is there some concept of stack where the 'return a' in try block is stored [if the example is altered, then it applies to try or catch block], and popped when control leaves the method1() only in the absence of a 'return' in finally block ?
Update: Solution concluded:
When the return statement is executed, the value to be returned is stored. When the finally block completes, that value is returned.
The finally block gets always executed as last. So the return in the finally block OVERWRITES the other returns in the try/catch blocks. It's a very bad practice to return or throw an exception from the finally block for this reason.
So to the original question - is it only in the absence of a 'return' in finally block ? NO. It doesn't matter.
Edit:
This question is asked on Apr 8 and is answered already. The question that is currently marked along with this as duplicate, is one asked at a later date [August 15]. Hence the new question is to be marked as duplicate, and not this one. However sharing the referenc eto a similar question is good.
A try block is executed before its finally block.
When the return statement is executed, the value to be returned is stored. When the finally block completes, that value is returned.
Note that a is not a value. a is a variable that stores a value. If you change a, you change a, you don't change the value that was stored for the return.
For the technical reason, we go to the Java Language Specification on the try-finally block
If execution of the try block completes abruptly for any other reason
R, then the finally block is executed, and 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).
A return statement completes abruptly, so if a try with a finally has a return and the finally also has a return, the finally's return supersedes the try's.
The finally block gets always executed as last. So the return in the finally block OVERWRITES the other returns in the try/catch blocks.
It's a very bad practice to return or throw an exception from the finally block for this reason.
Nothing to do with stacks: it's a matter of the fact that the finally block gets always executed as last block, so what is returned or thrown in the finally block erases and overwrites any previous exit value (returned object or exception thrown).
Try this:
try {
throw new NullPointerException();
}
finally {
return 0;
}
The return in the finally block will overwrite (swallow) the NullPointerException...
A try..catch..finally is actually compiled as a nested try..catch (of the exception you specify) within a try..catch (of "any"), and the "finally" block is cut & pasted by the compiler to the "end" of each code path. (I put "end" in quotes because I can't remember off the top of my head how it handles multiple exits from a "try" or "catch" block.) You can use the javap command (dis-assembler) to experiment on what byte code is generated by your version of Java to correspond to different try..catch..finally combinations.
I was responding to a slough of basic Java practice test questions and the correct answer to the following question slipped past me while taking the test.
Question: "If an exception is not caught, the finally block will run and the rest of the method is skipped - TRUE or FALSE?"
I am attempting to prove out the answer with the ThrowTest class (pasted at bottom) but I find Java exception handling to be somewhat unfamiliar. I compile the class as is with the ArrayIndexOutOfBoundsException portion of the catch block commented out. I then execute the class without passing an input parm thus creating an exception (the ArrayIndexOutOfBoundsException exception) . I would expect the final System.out.println would not execute and indeed it does not.
I find if I un-comment the ArrayIndexOutOfBoundsException catch block and recompile, the class then catches that specific error at run time and executes the "Rest of method" println at bottom. So, does my code prove the rest of "the rest of the method is skipped" or do I have to test it from a method other than the main method? Maybe there is a more straightforward method of testing it.
public class ThrowTest
{
public static void main(String[] args)
{
try
{
String anyString = args[0];
System.out.println("Try code executes");
}
catch(SecurityException e) //any old exception
{
System.err.println ("Error: SecurityException. ");
e.printStackTrace();
}
/* begin comment
catch(ArrayIndexOutOfBoundsException e)
{
System.err.println("Error: Caught ArrayIndexOutOfBoundsException. ");
e.printStackTrace();
}
end comment */
finally
{
System.out.println("finally block executes!");
}
System.out.println("Rest of method executes!");
}
}
Regardless of what exception is thrown a finally block will always execute (barring any strange situations like out of memory), so given the following
try {
// do some code
}
catch (SomeException e) {
// exception handling
}
finally {
doFinallyStuff();
}
doOtherStuff();
doFinallyStuff will always execute here. But doOtherStuff will only execute under the following conditions:
No exceptions are thrown
SomeException is thrown and the catch block does not rethrow the exception up the chain
If another exception is thrown in the try block that is not handled by the catch block, doFinallyStuff will still execute, but doOtherStuff will not
Your code is indeed proving that everything works as you described. When you catch an exception, then the rest of the method thereafter executes. If you let the method throw the exception, its execution halts immediately. Either way, the finally block will always execute.
I came up with this example while studying mock exams for OCPJP certification,
(from http://www.certpal.com version 1.6 exam part 3, Flow Control, Question number 8)
public class Oak
{
public static void main(String args[])
{
try
{
Integer i =null;
i.toString();
}
catch (Exception e)
{
try
{
System.out.println("One ");
Integer i =null;
i.toString();
}
catch (Exception x)
{
System.out.println("Two ");
Integer i =null;
i.toString();
}
finally
{
System.out.println("Three ");
Integer i =null;
i.toString();
}
}
finally
{
System.out.println("Four ");
}
}
}
I am fully aware that finally blocks always execute unless there is a System.exit(), so I traced the program and decided that it will have an output like this
One Two Exception in thread "main" java.lang.NullPointerException Three Exception in thread "main" java.lang.NullPointerException Four
However, it turns out the correct output is (according to the site and Eclipse debugging)
One Two Three Four Exception in thread "main" java.lang.NullPointerException
What I'm not understanding is, where did the exception which occurs in catch block with "Two" go?
There is no catch block underneath that, shouldn't it be thrown by the main thread too? Because it's not catched?
That exception is effectively lost - if a finally block throws an exception, any exception which was in the course of propagating becomes irrelevant, and the result of the finally block is the new exception.
Likewise if you return in a finally block, then that overall try/catch/finally will never throw an exception.
See section 14.20.2 of the Java Language Specification for details. In particular, various bits like this:
If the finally block completes abruptly for any reason, then the try statement completes abruptly for the same reason.
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.