Overriding Method with Exception - java

So here is the quote from the book:
The overriding method must NOT throw checked exceptions that are new or broader than those declared by the overridden method. For example, a method that declares a FileNotFoundException cannot be overridden by a method that declares a SQLException, Exception, or any other non-runtime exception unless it's a subclass of FileNotFoundException.
Now here is my question, if the method in the superclass throws an exception, then can the overriding method NOT throw an exception at all?
Because I just tried this in Java, where the overriding method did not throw any exceptions, and there was no error.
Please explain.

You can declare an overriding method as throwing less types of exceptions than the superclass, you just can't introduce new ones. The subclass method has to be compatible with the behavior of the superclass method. More exactly, you have to be able to substitute objects of the subclass for objects of the superclass without breaking anything (where adding a new checked exception to the throws clause would mean things calling it would have to change their code to handle it).
(The idea behind this is the Liskov Substitution Principle: a program should be able to deal with objects at a high level without getting bogged down in details about everything's exact type. If subclasses can introduce changes that mean the program has to pick them out and handle them differently then it defeats the purpose of abstraction.)
So an overriding method can be declared as throwing no checked exceptions at all (by omitting the throws clause entirely), because that doesn't require changes to any callers.
There are examples in the JDK, such as in java.io, where the subclass can't possibly throw an exception declared by the super class (see the ByteArrayOutputStream close method). Here the close method could have had its throws clause removed, since it never throws IOException. (Maybe it was left on the chance someone would want to subclass it with a version that did throw IOException?)

Related

Why is throws part of the method signature

Why throws, on a method, is part of its signature? It seems strange to include it. Here is an example where it is in the way:
#Overide
public void foo() {
throw new UnsupportedOperationException();
}
If anyone were to see this method from the outside, they might try to use it without knowing that it is not supported. They would only learn it on trying to run the code.
However, if they could do something like this they would know by looking at the method that it is not supported and if UnsupportedOperationException was not extending RuntimeException, they would get a compilation error. EDIT1: But this is not possible because throws is part of the signature so override will not work.
#Overide
public void foo() throws UnsupportedOperationException {
throw new UnsupportedOperationException();
}
This question concerns Java's design, so I know that it might be hard to answer without one of the people that work on it drops by and answers it, but I was hoping that maybe this question has been asked to them before or that there might be an obvious reason to have it this way to explain why.
The throws part does not indicate that the method is required to throw the mentioned exception(s), not even at particular occasions. It only tells that the function is allowed to do so.
Including throws UnsupportedOperationException will consequently not mean that the method is unsupported. Besides the UnsupportedOperationException is a RuntimeException so a method may throw that anyway.
Now for the reason one would require it in the signature of the method, it boils down to the ability to have checked exceptions at all. For the compiler to be able to decide if a method can only throw the specified exceptions it must be able to decide that the methods that it calls can't throw uncaught exceptions.
This means for example that overriding a method means that you can't add exceptions that might be thrown, otherwise you would break the possibility to verify that a method that calls that method can't throw anything else than it has specified. The other way around would be possible (but I'm not sure if Java supports that), overriding a method that may throw with one that may not throw.
So for example:
class B {
int fubar(int) throws ExceptionA {
}
int frob(int) throws ExceptionA {
return fubar(int);
}
}
class D extends B {
int fubar(int) throws ExceptionB {
}
}
Now frob is specified to possibly throw only ExceptionA, but in calling this.fubar it would open the possibility that something else is thrown, but fubar is defined to possibly only throw ExceptionA. That's why the D.fubar is an invalid override since that would open up the possibility that this.fubar actually throws ExceptionB and the compiler wouldn't be able to guarantee that frob doesn't throw ExceptionB.
Java has two different types of exceptions: checked Exceptions and
unchecked Exceptions.
Unchecked exceptions are subclasses of RuntimeException and you don't have to add a throws declaration. All other exceptions have to be handled in the method body, either with a try/catch statement or with a throws declaration.
Example for unchecked exceptions: IllegalArgumentException that is used sometimes to notify, that a method has been called with illegal arguments. No throws needed.
Example for checked exceptions: IOException that some methods from the java.io package might throw. Either use a try/catch or add throws IOException to the method declaration and delegate exception handling to the method caller.
One thing which no one has mentioned is a very important answer to your question:
Why throws, on a method, is part of its signature?
that throws, is NOT part of the method signature.
JLS 8.4.2. Method Signature makes it quite clear, that:
Two methods or constructors, M and N, have the same signature if they have the same name, the same type parameters (if any) (ยง8.4.4), and, after adapting the formal parameter types of N to the type parameters of M, the same formal parameter types.
If anyone were to see this method from the outside, they might try to use it without knowing that it is not supported.
That's exactly the main reason, why throws is designed in a way to be checked at compile-time for all Checked Exceptions - clients of the method will be aware what this method may possibly throw.
Checked Exceptions are enforced to be either handled or specified to be thrown (but this specification is NOT part of the method signature).
Unchecked Exceptions do not have to be either handled or specified to be thrown, and Oracle's tutorial are good at explaining - why:
Runtime exceptions represent problems that are the result of a programming problem, and as such, the API client code cannot reasonably be expected to recover from them or to handle them in any way. Such problems include arithmetic exceptions, such as dividing by zero; pointer exceptions, such as trying to access an object through a null reference; and indexing exceptions, such as attempting to access an array element through an index that is too large or too small.

Should I declare all exceptions thrown from a method in the method signature or just the super class of the exceptions?

When I throw checked exceptions from a method should I just declare the super class of the exceptions in the method signature or all the different types? If I have the following exceptions:
private class SuperException extends Exception {
}
private class SubExceptionOne extends SuperException {
}
private class SubExceptionTwo extends SuperException {
}
Should the method signature be:
void confirmAccount() throws SubExceptionOne, SubExceptionTwo;
or
void confirmAccount() throws SuperException;
In the last method signature, how do I tell other developers what exceptions that could be thrown from the method? If the different sub types need different handling?
The interface should be as stable as possible. So probably Super. Many libraries use the "Super" strategy, because exception specs cause far more annoyance in maintainability than readability or safety they add. Even the IOException is a Super that nearly all Java library code uses instead of declaring more specific exceptions. (But when they do declare more specific exceptions, it's because the contract is that more general IOExceptions won't be thrown. Read on.)
You might list Sub1 and Sub2 if you really do want to say each of those exceptions can be thrown, but don't want to say that any derivative of Super can be thrown. Perhaps Sub1 is NumberCrunchException and your method calls crunchNumbers() and users of your method can be assured that's the only exception-ful thing your method does. In that case the specific strategy is better.
If the different sub types need different handling, then definitely declare the two different exceptions. Never expect the developer using your method to guess that you are actually throwing different types of exceptions.
If you declare two distinct exceptions, and the user knows from the Javadoc that they are actually descendents of the same class, the user may choose to catch them both with a catch (SuperException e) rather than two individual catch clauses. But it depends on the user's choice.
If you don't declare them separately, your IDE is not going to add the appropriate #Throws to your Javadoc comment. And your Javadoc will therefore only indicate that you're throwing SuperException, which will leave the user in the dark. Solving this by just putting it in the text of the comment is not a real solution. If any tool is using reflection to determine what your method throws, it will not see the individual exceptions in the array returned from Method.getExceptionTypes().
If the functionality expected of the different exceptions is more or less the same and it's just a matter of how they will appear in the logs, it may be better to just use the parent exception, with different messages.
The throws clause is there to convey useful information to the calling method about what might go wrong during invocation of this method. That means that how specific you are will depend on how much information you want to convey; and that will be application-dependent.
For instance, declaring throws Exception is almost always a bad idea: the information this conveys is just "something might go wrong", which is too vague to be useful. But whether calling classes are going to need perfectly fine-grained information in the throws clause is something you need to decide by looking at your program. There's no set answer.

Constructor publishing itself then throwing exception [duplicate]

Are constructors allowed to throw exceptions?
Yes, constructors can throw exceptions. Usually this means that the new object is immediately eligible for garbage collection (although it may not be collected for some time, of course). It's possible for the "half-constructed" object to stick around though, if it's made itself visible earlier in the constructor (e.g. by assigning a static field, or adding itself to a collection).
One thing to be careful of about throwing exceptions in the constructor: because the caller (usually) will have no way of using the new object, the constructor ought to be careful to avoid acquiring unmanaged resources (file handles etc) and then throwing an exception without releasing them. For example, if the constructor tries to open a FileInputStream and a FileOutputStream, and the first succeeds but the second fails, you should try to close the first stream. This becomes harder if it's a subclass constructor which throws the exception, of course... it all becomes a bit tricky. It's not a problem very often, but it's worth considering.
Yes, they can throw exceptions. If so, they will only be partially initialized and if non-final, subject to attack.
The following is from the Secure Coding Guidelines 2.0.
Partially initialized instances of a non-final class can be accessed via a finalizer attack. The attacker overrides the protected finalize method in a subclass, and attempts to create a new instance of that subclass. This attempt fails (in the above example, the SecurityManager check in ClassLoader's constructor throws a security exception), but the attacker simply ignores any exception and waits for the virtual machine to perform finalization on the partially initialized object. When that occurs the malicious finalize method implementation is invoked, giving the attacker access to this, a reference to the object being finalized. Although the object is only partially initialized, the attacker can still invoke methods on it (thereby circumventing the SecurityManager check).
Absolutely.
If the constructor doesn't receive valid input, or can't construct the object in a valid manner, it has no other option but to throw an exception and alert its caller.
Yes, it can throw an exception and you can declare that in the signature of the constructor too as shown in the example below:
public class ConstructorTest
{
public ConstructorTest() throws InterruptedException
{
System.out.println("Preparing object....");
Thread.sleep(1000);
System.out.println("Object ready");
}
public static void main(String ... args)
{
try
{
ConstructorTest test = new ConstructorTest();
}
catch (InterruptedException e)
{
System.out.println("Got interrupted...");
}
}
}
Yes, constructors are allowed to throw exceptions.
However, be very wise in choosing what exceptions they should be - checked exceptions or unchecked. Unchecked exceptions are basically subclasses of RuntimeException.
In almost all cases (I could not come up with an exception to this case), you'll need to throw a checked exception. The reason being that unchecked exceptions (like NullPointerException) are normally due to programming errors (like not validating inputs sufficiently).
The advantage that a checked exception offers is that the programmer is forced to catch the exception in his instantiation code, and thereby realizes that there can be a failure to create the object instance. Of course, only a code review will catch the poor programming practice of swallowing an exception.
Yes.
Constructors are nothing more than special methods, and can throw exceptions like any other method.

Why do I need "throws Exception"?

This is my code:
public static void main(String[] args) throws Exception {
Scanner s = new Scanner(new File("story.txt"));
someFunction(s);
}
The story.txt is in the the project root, but without the throws Exception the code doesn't run.
Why?
You don't need throws Exception, but it won't hurt any if you leave it like that.
What you do need is a throws FileNotFoundException because the Scanner constructor you are using is declared to throw that exception, and you don't catch any exceptions.
You should throw a FileNotFoundException, because the Scanner constructor can throw one. Since it is a checked exception, you must either pass it further up the stack or handle it somehow.
I asked my professor, and he explained that Java makes you handle exceptions.
You must be ready for them if a function might throw them.
You can either use:
try {...
} catch { ...
}
to handle exceptions yourself, or you can use:
throws Exceptions
to "pass them to the next level" (handle them later).
Thanks for the help!
If a method or constructor either throws an exception which isn't derived from RuntimeException, or calls any method whose declaration indicates it must do so, then it must either catch that exception or declare that the exception may be thrown from it. This language feature is almost a good thing, but in its present form has been widely acknowledged as a mistake. Nonetheless, it is sufficiently firmly established as part of the Java language that it's not going anywhere.
Conceptually, it's good to make a distinction between exceptions which are likely to have special meaning to a method's immediate caller and those which would not. Unfortunately, the fact that an exception would have a special meaning to its immediate caller doesn't mean the immediate caller is going to be interested in it. The way Java implements checked exceptions, if Foo is declared as throwing a checked exception and Bar calls Foo but isn't prepared to handle that exception, it must either declare itself as throwing that exception (even though it's unlikely to have any special meaning to its caller), or else catch the exception it has no hope of handling (perhaps rethrowing as a type derived from RuntimeException). Catching and rethrowing as a type derived from RuntimeException is semantically the best approach, but ends up being the most verbose, especially if one wants to avoid making RuntimeException objects.
Note also that because there's no way a chained constructor call can be wrapped in a try block, any constructor which is chained through a constructor that is declared as throwing any exceptions is required to declare itself as throwing those same exceptions. Things are implemented that way for a reason, but it unfortunately makes it difficult for a class constructor to distinguish exceptions which got thrown as part of the base-class construction process and those which got thrown in the construction of the derived class.
Because, Scanner s = new Scanner(new File("story.txt")); throws checked FileNotFoundException. You should throws or catch checked exception for compilation.

Java constructors [duplicate]

Are constructors allowed to throw exceptions?
Yes, constructors can throw exceptions. Usually this means that the new object is immediately eligible for garbage collection (although it may not be collected for some time, of course). It's possible for the "half-constructed" object to stick around though, if it's made itself visible earlier in the constructor (e.g. by assigning a static field, or adding itself to a collection).
One thing to be careful of about throwing exceptions in the constructor: because the caller (usually) will have no way of using the new object, the constructor ought to be careful to avoid acquiring unmanaged resources (file handles etc) and then throwing an exception without releasing them. For example, if the constructor tries to open a FileInputStream and a FileOutputStream, and the first succeeds but the second fails, you should try to close the first stream. This becomes harder if it's a subclass constructor which throws the exception, of course... it all becomes a bit tricky. It's not a problem very often, but it's worth considering.
Yes, they can throw exceptions. If so, they will only be partially initialized and if non-final, subject to attack.
The following is from the Secure Coding Guidelines 2.0.
Partially initialized instances of a non-final class can be accessed via a finalizer attack. The attacker overrides the protected finalize method in a subclass, and attempts to create a new instance of that subclass. This attempt fails (in the above example, the SecurityManager check in ClassLoader's constructor throws a security exception), but the attacker simply ignores any exception and waits for the virtual machine to perform finalization on the partially initialized object. When that occurs the malicious finalize method implementation is invoked, giving the attacker access to this, a reference to the object being finalized. Although the object is only partially initialized, the attacker can still invoke methods on it (thereby circumventing the SecurityManager check).
Absolutely.
If the constructor doesn't receive valid input, or can't construct the object in a valid manner, it has no other option but to throw an exception and alert its caller.
Yes, it can throw an exception and you can declare that in the signature of the constructor too as shown in the example below:
public class ConstructorTest
{
public ConstructorTest() throws InterruptedException
{
System.out.println("Preparing object....");
Thread.sleep(1000);
System.out.println("Object ready");
}
public static void main(String ... args)
{
try
{
ConstructorTest test = new ConstructorTest();
}
catch (InterruptedException e)
{
System.out.println("Got interrupted...");
}
}
}
Yes, constructors are allowed to throw exceptions.
However, be very wise in choosing what exceptions they should be - checked exceptions or unchecked. Unchecked exceptions are basically subclasses of RuntimeException.
In almost all cases (I could not come up with an exception to this case), you'll need to throw a checked exception. The reason being that unchecked exceptions (like NullPointerException) are normally due to programming errors (like not validating inputs sufficiently).
The advantage that a checked exception offers is that the programmer is forced to catch the exception in his instantiation code, and thereby realizes that there can be a failure to create the object instance. Of course, only a code review will catch the poor programming practice of swallowing an exception.
Yes.
Constructors are nothing more than special methods, and can throw exceptions like any other method.

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