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How to handle exceptions that you didn't expect even thought it is declared on the documentation?

When there are methods that throw exceptions and you know these exceptions will not be thrown, what should you do?
Many times I see people just logging the exception, but I wonder if there is a build in exception in java that means something like: "This exception should not have been thrown".
For example, imagine I have a code that calls StaticClass.method(someObject) and this method throws a SpecificException when someObject is not valid. What should you do in the catch block?
try {
StaticClass.method(someobject);
} catch (SpecificException e) {
// what should I do here?
}

If when calling the method you know for sure that it will not throw an exception because of previous checks you should throw a RuntimeException wrapping the SpecificException.
try {
StaticClass.method(someobject);
} catch (SpecificException e) {
//This is unexpected and should never happen.
throw new RuntimeException("Error occured", e);
}
Some methods already throw a RuntimeException when they fail to perform their purpose.
//Here we know for sure that parseInt(..) will not throw an exception so it
//is safe to not catch the RuntimeException.
String s = "1";
int i = Integer.parseInt(s);
//Here instead parseInt(..) will throw a IllegalArgumentException which is a
//RuntimeException because h is not a number. This is something that should
//be fixed in code.
s = "h";
i = Integer.parseInt(s);
RuntimeExceptions don't require a try/catch block and the compiler will not be mad at you for not catch them. Usually they are thrown where something in your app code is wrong and should be fixed. Anyway there are cases where catching a RuntimeException is useful.

Can a checked exception be used to throw an unchecked exception?

I am running the following code:
import java.io.FileNotFoundException;
import java.io.IOException;
class Basic1 {
int c;
void calculation(int a, int b) throws Exception {
c = a / b;
}
}
class Basic extends Basic1 {
void calculation(int a, int b) throws IOException {
c = a / b;
RuntimeException ae = new ArithmeticException();
throw ae;
}
public static void main(String[] args) {
int a = 10;
int b = 0;
int c;
Basic ba = new Basic();
try {
ba.calculation(a, b);
} catch (IOException e) {
System.out.println("Zero can't be there in the denominator. : IoException");
} catch (ArithmeticException e) {
System.out.println("Zero can't be there in the denominator. : Arthimetic Exception");
} catch (Exception e) {
System.out.println("Zero can't be there in the denominator. : Exception");
}
}
}
The program is compiling successfully and outputting "Zero can't be there in the denominator. : Arthimetic Exception" (output is as expected).
My question is how is the program able to compile successfully? Why am I not getting an error when I throw an IOException while inside calculation() I am creating a RuntimeException object?
My second question is given that the program enters the catch (ArithmeticException e) clause, is the compiler deciding at run time which catch will execute? Do I understand correctly?

For your first question, methods using the throws E (where E is some arbitrary checked exception) construct don't have to throw anything. All it means is that anyone calling your method has to be able to handle E, which allows your method to throw E without handling it. This would otherwise be a compilation error. Usually, the documentation of your method specifies when E is thrown. In the words of the JLS:
Essentially, for each checked exception that can result from execution of the body of a method or constructor, a compile-time error occurs unless its exception type or a supertype of its exception type is mentioned in a throws clause in the declaration of the method or constructor.
This doesn't say you must throw the given checked exception within the method body. Note that you can use throws with unchecked exceptions, but this has no effect on the program, as unchecked exceptions may be thrown anywhere without being handled. That's the point of them.
In your code, you throw a RuntimeException. Any class extending RuntimeException (including itself) is unchecked by definition. ArithmeticException is one of these, so you may assign it to a variable of type RuntimeException and then throw it without specifying it in the throws clause.
For your second question, I'll assume that the compiler doesn't optimise away the various catch clauses. In that case, no, the catch which is executed is not determined at compile time. When any exception is thrown in your program, it propagates up the stack until it is caught. If it isn't, it halts execution and the stack trace is printed to System.err. Since an ArithmeticException is thrown, the first catch clause does not match but the second one does, so is executed. The third one (catch (Exception e)) would match, but order matters with catch clauses and only one is executed, so that clause is not executed here.
However, in this case, it is not inconceivable that the compiler optimises away your catch clauses and directs the program straight to the code in the catch (ArithmeticException e) clause, as your method will always throw ArithmeticException. It must be considered, however, that any line of code may throw some obscure error like OutOfMemoryError; the compiler must allow for such edge cases when optimising.

Exception hierarchy/try-multi-catch

try {
throw new FileNotFoundException();
} catch (IOException e) {
e.printStackTrace();
}
catch (Exception e) {
e.printStackTrace();
}
Can someone tell me why the second catch block is not considered as unreachable code by the compiler? But in the following case:
try {
throw new FileNotFoundException();
} catch (Exception e) {
e.printStackTrace();
}
catch (IOException e) {
e.printStackTrace();
}
The second catch block is considered unreachable?
After all, FileNotFoundException comes under IOException, just as it comes under Exception.
Edit Please clarify:
The compiler will know an exception is thrown by a method based on the method's throws clause. But it may not necessarily know the specific type of exception (under that class of exception). So if a method throws exception 'A', compiler won't know if the actual exception is 'A' or a subtype of 'A' because this is only determined at runtime. The compiler however will know that an exception of type 'X' is never thrown, so giving a catch block for X is erroneous. Is this right?

The compiler cannot assume that the only possible exception thrown from your try block will be a FileNotFoundException. That's why it doesn't consider the 2nd catch block to be unreachable in your first code sample.
What if, for some unknown reason, a RuntimeException gets thrown while creating the FileNotFoundException instance (entirely possible)? What then?
In your first code sample, that unexpected runtime exception would get caught by the 2nd catch block, while the 1st block would take care of the FileNotFoundException if it gets thrown.
However, in your 2nd code sample, any and all exceptions would get caught by the 1st catch block, making the 2nd block unreachable.
EDIT:
To better understand why the catch(Exception e) block in your first code is not considered unreachable by the compiler, try the following code, and notice how the the 2nd catch is definitely reachable:
public class CustomIOException extends IOException {
public CustomIOException(boolean fail) {
if (fail) {
throw new RuntimeException("the compiler will never know about me");
}
}
}
public static void main(String[] args) {
try {
throw new CustomIOException(true);
} catch(IOException e) {
System.out.println("Caught some IO exception: " + e.getMessage());
} catch(Exception e) {
System.out.println("Caught other exception: " + e.getMessage());
}
}
Output:
Caught other exception: the compiler will never know about me

TL;DR
The compiler considers that FileNotFoundException() may not be the only Exception thrown.
Explaination
JLS§11.2.3 Exception Checking
A Java compiler is encouraged to issue a warning if a catch clause can
catch (§11.2) checked exception class E1 and the try block
corresponding to the catch clause can throw checked exception class
E2, a subclass of E1, and a preceding catch clause of the immediately
enclosing try statement can catch checked exception class E3 where E2
<: E3 <: E1.
That means that if the compiler considers that the only exception possibly thrown by your catch block is a FileNotFoundException(), it will warn you about your second catch block. Which is not the case here.
However, the following code
try{
throw new FileNotFoundException();
} catch (FileNotFoundException e){
e.printStackTrace();
} catch (IOException e){ // The compiler warns that all the Exceptions possibly
// catched by IOException are already catched even though
// an IOException is not necessarily a FNFException
e.printStackTrace();
} catch (Exception e){
e.printStackTrace();
}
This happens because the compiler evaluates the try block to determine which exceptions has the possibility to be thrown.
As the compiler does not warn us on Èxception e, it considers that other exceptions may be thrown (e.g RunTimeException). Since it is not the compiler's work to handle those RunTimeExceptions, it lets it slip.
Rest of the answer is intersting to read to understand the mechanism behind exception-catching.
Schema
As you may see, Exception is high in the hierarchy so it has to be declared last after IOException that is lower in the hierarchy.
Example
Imagine having an IOException thrown. As it is inherited from Exception, we can say IOException IS-A Exception and so, it will always be catched within the Exception block and the IOException block will be unreachable.
Real Life Example
Let's say, you're at a store and have to choose pants. The seller tells you that you have to try the pants from the largest ones to the smallest ones and if you find one that you can wear (even if it is not your size) you must take it.
You'll find yourself buying pants too large for your size and you'll not have the chance to find the pants that fits you.
You go to another store : there, you have the exact opposite happening. You can choose your pants from smallest to largest and if you find one you can wear, you must take it.
You'll find yourself buying pants at your exact size.
That's a little analogy, a bit odd but it speaks for itself.
Since Java 7 : multi-catch
Since Java 7, you have the option to include all the types of Exceptions possibly thrown by your try block inside one and only catch block.
WARNING : You also have to respect the hierarchy, but this time, from left to right.
In your case, it would be
try{
//doStuff
}catch(IOException | Exception e){
e.printStackTrace();
}
The following example, which is valid in Java SE 7 and later,
eliminates the duplicated code:
catch (IOException|SQLException ex) {
logger.log(ex);
throw ex;
}
The catch clause specifies the types of exceptions that the block can
handle, and each exception type is separated with a vertical bar (|).

First case:
catch (IOException e) { // A specific Exception
e.printStackTrace();
}
catch (Exception e) { // If there's any other exception, move here
e.printStackTrace();
}
As you can see, first IOException is catched. This means we are aiming at just one specific exception. Then in second catch, we aim for any other Exceptions other than IOException. Hence its logical.
In second:
catch (Exception e) { // Move here no matter whatever exception
e.printStackTrace();
}
catch (IOException e) { // The block above already handles *Every exception, hence this won't be reached.
e.printStackTrace();
}
We catched any exception (whether its IOException or some other Exception), right in the first block. Hence second block will not be reached because everything is already included in first block.
In other words, in first case, we aim at some specific exception, than at any other exceptions. While in second case, we first aim at all/any Exception, than at a specific exception. And since we already dealt will all exceptions, having a specific exception later won't make any logical sense.

Why does an empty try block with catch Exception compile? [duplicate]

This question already has answers here:
Java unreachable catch block compiler error
(7 answers)
Closed 7 years ago.
A try block without any code :
try {
} catch (Exception ex) {
// what Exception it is catching
ex.printStackTrace();
}
The absence of any code means that throwing an exception is impossible, so why doesn't this give an "unreachable catch block" compile error?

Exception includes RuntimeExceptions, which are unchecked and don't need to be declared, so Exception can always be validly caught.
I think this is an unimportant edge case.

That's valid Java syntax. It's the same as having an empty if-block:
if (condition) {
}
... or defining an empty method:
public void empty() {
}
... or only having comments as part of the body:
try {
// try body
} catch (Exception e) {
// catch body
}
All of that is valid syntax, so the compiler is happy. Further, since a blank line / empty body is totally ok, no exceptions will be thrown in the body of the try block during runtime, so the code would execute just fine as well.

So I am assuming that your question is what Exception will get caught and the answer is none. The exception would only be caught if while running the code within the try block throws an exception. It will then check to see if that exception was caught (FYI Exception will catch all exceptions), if yes it handles it inside the catch block otherwise it causes an error. Since there is nothing in the try block the exception will never be caught because no exception can ever be thrown.

Non throwable code inside of a try/catch performance java

Will there be any negative performance implications on the following code due to methods that do not throw exceptions being inside of a try block?
String user;
try {
user = SomeHelper.getUserName(); //Does not throw anything explicitly
if (SomeHelper.isSomething(user)) { //Does not throw anything explicitly
user.doSomeSafeThing(this); //Does not throw anything explicitly
}
else {
user.doOtherSafeThing(this); //Does not throw anything explicitly
}
SomeOtherHelper.methodThatCanBlow(User.getCredentials(context)); //This will throw exception
} catch (Exception e) {
e.printStackTrace();
}
I believe that the compiler will not refactor any of this code due to the fact that any of those methods could potentially throw a NullPointerException, or another RuntimeException and actually be caught in the catch block.
So I beg to ask the question, would the following code be more efficient?
String user;
user = SomeHelper.getUserName(); //Does not throw anything explicitly
if (SomeHelper.isSomething(user)) { //Does not throw anything explicitly
user.doSomeSafeThing(this); //Does not throw anything explicitly
}
else {
user.doOtherSafeThing(this); //Does not throw anything explicitly
}
try {
SomeOtherHelper.methodThatCanBlow(User.getCredentials(context)); //This will throw exception
} catch (Exception e) {
e.printStackTrace();
}

Will there be any negative performance implications on the following code due to methods that do not throw exceptions being inside of a try block?
No, it won't have any performance implications, but:
I would generally put minimal code in a try block anyway
I would try to avoid catching Exception, instead catching specific exceptions
Just printing a stack trace is very rarely the right way of "handling" and exception that you catch. Usually you want to abort the whole operation, whatever that consists of