I have a problem understanding how the try{} catch(Exception e){...} works!
Let's say I have the following:
try
{
while(true)
{
coord = (Coordinate)is.readObject();//reading data from an input stream
}
}
catch(Exception e)
{
try{
is.close();
socket.close();
}
catch(Exception e1)
{
e1.printStackTrace();
}
}
Section 2
try
{
is.close();
db.close();
}
catch(Exception e)
{
e.printStackTrace();
}
Let's say my while() loop throws an error because of an exception of is stream.
This will get me out of the infinite loop and throw me in the first catch(){............}
block.
My question is the following:
After throwing an exception, getting out of the loop while() and reaching to
catch(){
}
Will my program continue his execution and move on to section 2? As long as the exception was caught? Or everything ends in the first catch()?
I think you want to use finally after your first catch [catch (Exception e)] to close your streams:
try {
// Do foo with is and db
} catch (Exception e) {
// Do bar for exception handling
} finally {
try {
is.close();
db.close();
} catch (Exception e2) {
// gah!
}
}
As long as no exceptions are thrown in your catch clause, your program will continue execution after your catch (or finally) clause. If you need to rethrow the exception from the catch clause, use throw; or throw new Exception(ex). Do not use throw ex, as this will alter the stack trace property of your exception.
After the exception is caught, execution continues after the try/catch block. In this case, that's your section 2.
An uncaught exception will terminate the thread, which might terminate the process.
Yes, you are right. It will move to Section 2.
If you want your Section 2 bound to happen, irrespective of any exception generated, you might want to enclose them in a finally block.
try {
// Do foo with is and db
}
catch (Exception e) {
// Do bar for exception handling
}
// Bound to execute irrespective of exception generated or not ....
finally {
try {
is.close();
db.close();
}
catch (Exception e2) {
// Exception description ....
}
}
Related
Basically i have 2 different commands i could possibly execute. If the first one does not work, I want to execute the second command.
Is there some easier, cleaner way to do this?
What would i even put in the if statement?
try
{
// code that may throw an exception
driver.findElement(By.id("component-unique-id-31")).click();
}
catch (Exception ex)
{
// Print to console
}
if(previous command threw an exception){
try
{
//Another command i want executed if the above fails
driver.findElement(By.xpath("//h3[normalize-space()='Something']")).click();
}
catch (Exception ex)
{
// handle
}
}
You can put the second command inside the catch block of the first try-catch, as following:
try
{
// code that may throw an exception
driver.findElement(By.id("component-unique-id-31")).click();
}
catch (Exception ex1)
{
try
{
//Another command i want executed if the above fails
driver.findElement(By.xpath("//h3[normalize-space()='Something']")).click();
}
catch (Exception ex2)
{
// handle
}
}
You can wrap up both the try-catch{} blocks within a single try-catch-finally{} block as follows:
try {
new WebDriverWait(driver, 20).until(ExpectedConditions.elementToBeClickable(By.id("component-unique-id-31"))).click();
System.out.println("Within in try block.");
}
catch(TimeoutException e) {
new WebDriverWait(driver, 20).until(ExpectedConditions.elementToBeClickable(By.xpath("//h3[normalize-space()='Something']"))).click();
System.out.println("Within in catch block.");
} finally {
System.out.println("The 'try catch' is finished.");
}
PS: You must avoid catching the raw exception.
I want to execute my method callmethod if the condition inside the IF statement is met. Else it should execute the catch block. But during implementation, if the condition is not met, it does not go to the catch block.
try{
if(count==0)
callmethod();
}
catch (Exception e){
System.out.println(e);
}
This is a good application for methods:
try {
if (count == 0) {
callOneMethod();
}
else {
callOtherMethod();
}
catch (Exception e) {
callOtherMethod();
}
That way you don't have any duplicated code and you're not doing weird things with exceptions in non-exceptional cases.
Since you are trying to hit the catch block, you need to throw an exception if your parameter is not met (i.e. count != 0).
Example:
try {
if(count==0){
callmethod();
} else {
throw new SomeException("some message");
}
}
catch (Exception e){
System.out.println(e);
}
This question already has answers here:
Order of catching exceptions in Java
(3 answers)
Closed 6 years ago.
Here is my program.
try {
int a = 1/0;
}
catch(Exception e) {
system.out.println("Exception block"+e);
}
catch(ArithmeticException e) {
system.out.println("Inside ArithmeticException block");
}
finally {
system.out.println("Inside Finally block");
}
In the above program i have two catch blocks and one finally block.
Which catch block will execute? Because I define the parent catch block first. So it leads to an error? Can any one help me?
I assumed that "ArithmeticException and Finally block will be executed"
Catching Exception class will not give any error. However, it is not recommended.
In this case, Exception catch block will execute first and then finally block will execute.
If you want that your ArithmeticException block should execute, put this block before Exception catch block.
Update code -
try{
int a = 1/0;
}
catch(ArithmeticException e){
System.out.println("Inside ArithmeticException block");
}
catch(Exception e) {
System.out.println("Exception block"+e);
}
finally{
System.out.println("Inside Finally block");
}
Catching Exception class will not give any error.
In this case, Exception catch block will execute first and then finally block will execute.
If you want that your ArithmeticException block should execute, put this block before Exception catch block.
Here is Updated code
try {
int a = 1/0;
}
catch(Exception e) {
System.out.println("Exception block"+e);
}
finally {
System.out.println("Inside Finally block");
}
}
Not sure if this has already been answered, but.
I know that in java there is the try, catch and finally blocks, but is there one which is only called if try has no errors/exceptions?
Currently after stating the command that needs to be run, I'm setting a boolean to true, and after the try and catch block, the program checks for if the boolean is true.
I'm pretty sure that there is a much simpler way, help is appreciated!
Just put your code after the try...catch block and return in the catch:
boolean example() {
try {
//dostuff
} catch (Exception ex) {
return false;
}
return true;
}
This would also work if you put the return true at the end of the try block as the code would jump to the catch on error and not execute the rest of the try.
void example() {
try {
//do some stuff that may throw an exception
//do stuff that should only be done if no exception is thrown
} catch (Exception ex) {
throw new RuntimeException(ex);
}
}
No, there is no block called only if no exceptions were raised.
The catch block is called if there were exceptions, finally is called regardless.
As stated, you can emulate such a beast with something like:
bool completed = false;
try {
doSomeStuff();
completed = true;
} catch (Exception ex) {
handleException();
} finally {
regularFinallyHandling();
if (completed) {
thisIsTheThingYouWant();
}
}
but there's nothing built into the language itself that provides this functionality.
What happens if you throw an error in a finally block? Does it get handled in one of the corresponding catch clauses?
Only if you put another try-catch block in the finally block. Otherwise it's an error like any other.
You need to include try-catch blocks inside the finally or catch blocks.
e.g.:
try {
// your code here
} finally {
try {
// if the code in finally can throw another exception, you need to catch it inside it
} catch (Exception e) {
// probably not much to do besides telling why it failed
}
} catch (Exception e) {
try {
// your error handling routine here
} catch (Exception e) {
// probably not much to do besides telling why it failed
}
}
It will not handle exception until it is caught in finally block it self.
public static void main(String[] args) throws Exception {
try {
System.out.println("In try");
} catch (Exception e) {
System.out.println("In catch");
} finally{
throw new Exception();
}
}
Above code will throw exception but if you do as follows it will work:
public static void main(String[] args){
try {
System.out.println("In try");
} catch (Exception e) {
System.out.println("In catch");
} finally{
try{
throw new Exception();
}catch(Exception e){}
}
}
Nope. It would be caught by a catch where then entire try/catch/finally was nested within another try/catch. The exception would otherwise be thrown out of the function, and would be handled by the caller of the function.
No it doesn't. You will have to handle with it IN the finally block or define a proper throw declaration in the method description.
No, a catch block can only catch exceptions thrown within the corresponding try block - not a finally block. (Of course, if that finally block is within another try block, the catch clauses for that try block are still used.)
The relevant section in the JLS is 14.20.2. Each of the flows listed there has something like this:
If the finally block completes abruptly for any reason, then the try statement completes abruptly for the same reason.
In other words, there's no attempt for any catch clauses associated with the finally block to handle the exception.
The order of execution is normally directly indicated by the order of statements: 1. try, 2. catch exceptions in the specified order (only one catch is executed), 3. finally.
So when the finally block is executed (note that this is always the case, even in the case of a return statement or exception being thrown in the try or catch blocks) the execution of try statement is in its last phase and thus it cannot catch further throwables. As already pointed out, the exception has to be handled in a location further down the stack (or up, depends on the view point ;) ).