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:
System.exit();try or catch block;thread.stop() does not necessarily prevent finally block from being executed. - Piotr Findeisenfinally block will be called after the try block, and before control passes to the following statements. That's consistent with the try block involving an infinite loop and hence the finally block never actually being invoked. - Andrzej Doyle
Example code:
public static void main(String[] args) {
System.out.println(Test.test());
}
public static int test() {
try {
return 0;
}
finally {
System.out.println("finally trumps return.");
}
}
Output:
finally trumps return.
0
finally-clause with return 2; is not allowed (Compiler-Error). - Alexander Pacha
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
trystatement with afinallyblock is executed by first executing thetryblock. Then there is a choice:
- If execution of the
tryblock completes normally, [...]- If execution of the
tryblock completes abruptly because of athrowof a value V, [...]- If execution of the
tryblock completes abruptly for any other reason R, then thefinallyblock is executed. Then there is a choice:
- If the finally block completes normally, then the
trystatement completes abruptly for reason R.- If the
finallyblock completes abruptly for reason S, then thetrystatement 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
returnstatement with noExpressionattempts to transfer control to the invoker of the method or constructor that contains it.A
returnstatement with anExpressionattempts to transfer control to the invoker of the method that contains it; the value of theExpressionbecomes the value of the method invocation.The preceding descriptions say "attempts to transfer control" rather than just "transfers control" because if there are any
trystatements within the method or constructor whosetryblocks contain thereturnstatement, then anyfinallyclauses of thosetrystatements will be executed, in order, innermost to outermost, before control is transferred to the invoker of the method or constructor. Abrupt completion of afinallyclause can disrupt the transfer of control initiated by areturnstatement.
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();
}
}
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.
i was not a primitive, but an Integer object. - Yamcha
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");
}
}
Output:
X
finally trumps return... sort of
0
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:
Also a return in finally will throw away any exception. http://jamesjava.blogspot.com/2006/03/dont-return-in-finally-clause.html
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");
}
}
}
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.
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;
}
}
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.
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
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.
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.
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.
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());
}
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-
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;
}
}
sb.setLength(0) in finally? - user7294900
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.
finally at the first place... - glglgl
Because the final is always be called in whatever cases you have. You don't have exception, it is still called, catch exception, it is still called
Consider this in a normal course of execution (i.e without any Exception being thrown): if method is not 'void' then it always explicitly returns something, yet, finally always gets executed
If an exception is thrown, finally runs. If an exception is not thrown, finally runs. If the exception is caught, finally runs. If the exception is not caught, finally runs.
Only time it does not run is when JVM exits.
Finally block always execute whether exception handle or not .if any exception occurred before try block then finally block will not execute.
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.
Yes, because no control statement can prevent finally from being executed.
Here is a reference example, where all code blocks will be executed:
| x | Current result | Code
|---|----------------|------ - - -
| | |
| | | public static int finallyTest() {
| 3 | | int x = 3;
| | | try {
| | | try {
| 4 | | x++;
| 4 | return 4 | return x;
| | | } finally {
| 3 | | x--;
| 3 | throw | throw new RuntimeException("Ahh!");
| | | }
| | | } catch (RuntimeException e) {
| 4 | return 4 | return ++x;
| | | } finally {
| 3 | | x--;
| | | }
| | | }
| | |
|---|----------------|------ - - -
| | Result: 4 |
In the variant below, return x; will be skipped. Result is still 4:
public static int finallyTest() {
int x = 3;
try {
try {
x++;
if (true) throw new RuntimeException("Ahh!");
return x; // skipped
} finally {
x--;
}
} catch (RuntimeException e) {
return ++x;
} finally {
x--;
}
}
References, of course, track their status. This example returns a reference with value = 4:
static class IntRef { public int value; }
public static IntRef finallyTest() {
IntRef x = new IntRef();
x.value = 3;
try {
return x;
} finally {
x.value++; // will be tracked even after return
}
}
finally; finalizer == thefinalize()method. - jaco0646