Consider the following example:
public class LambdaArgsTest {
private static void display(Supplier<?> arg) {
try {
// this is the place where the Exception("wrong") might be thrown
// and it is in fact handled
System.out.println(arg.get());
} catch (Exception e) {
e.printStackTrace();
}
}
public static void main(String[] args) {
display(() -> {
if(/*some condition*/) {
// this statement will be rejected due to unhandled exception
throw new Exception("wrong");
}
return "abcde";
});
}
}
Here comes the question: the lambda argument in the above example is an object that'll be executed later inside the "display()" method. It is obviously not executed when passing the arguments to the "display()".
Why is it rejected by the compiler? I think it's quite reasonable to surround it by try...catch only when the lambda is actually called.
This is because of the signature of Supplier functional interface:
T get();
As you see, the method get is not declared to throw Exception (neither any other checked exception).
In Java, there are checked vs unchecked exceptions (unchecked exceptions are those which inherit from RuntimeException). Checked exceptions must be handled, either by catching them in a catch block, or by declaring that the method throws that exception.
If the signature of Supplier.get were:
T get() throws Exception:
The code would compile fine.
Try throwing RuntimeException instead of Exception and the code will compile fine.
EDIT: As per Peter Lawrey's suggestion in the comments, if you do need to throw a checked exception from within your lambda expression, you could use e.g. Callable, whose only one method's signature is as follows:
T call() throws Exception;
You'd just need to pass a Callable to your display method instead of a Supplier.
Related
I understand that catching a checked exception that is never thrown in the corresponding try block is invalid. Because the compiler itself would have forced the programmer to handle the exception if it was to occur.
For example, this code snippet -
try
{
}
catch(IOException e)
{
}
is invalid.
But why doesn't the compiler work the same way for the method that throws a checked exception that has never been thrown in the body of the method?
For example, this code snippet -
void test() throws IOException
{
}
is surprisingly valid.
Please explain the reason behind it. TIA.
Because you might want to allow subclasses to throw this exception.
public abstract class Parent {
public void doStuff() {}
}
public class Child {
// this is illegal, because you throw more exceptions than the overridden method
public void doStuff() throws IOException {}
}
This is even more intuitive for interfaces, the (non-default) methods of which will never throw an exception but can declare it.
It is a compile-time error if a catch clause can catch checked
exception class E1 and it is not the case that the try block
corresponding to the catch clause can throw a checked exception class
that is a subclass or superclass of E1, unless E1 is Exception or a
superclass of Exception.
This tells you all these are valid:
try { }
catch(Exception e){}
--
try{ }
catch(NullPointerException e) {}
--
try{ }
catch(ArrayIndexOutOfBoundsException e) {}
--
try{ }
catch(RuntimeException e) {}
--
try{ }
catch(Error e) {}
--
try{ }
catch(Throwable e){ }
It's not related to a void block. It's related to an impossible path to your checked subclass of Exception.
What happens with subclasses of Exception? For example IOexception: Those are unreachable catch blocks. Nothing in the try block can lead to that exception, so the compiler just tells you: this block would never be executed, as it would never catch that subException.
The difference with throws: The concept of unreachable code doesn't exist in this context. The possibility of a method throwing an exception doesn't end in the deffinition of the method. For example:
abstract void readFile(String path) throws IOException;
This method doesn't even have a block, as it's an abstract method. It's easy to guess that this line won't ever throw any IOException. But it defines a behaviour for the extensions that will implement it. In order to override it, your method must throw the IOException.
In the same way, if someone overrides your test method:
#Override
void test() throws IOException
{
readFile(file);
}
It's not impossible to happen, in contrary to your first try-catch block.
Java follows the rule "handle or declare".
If there is a checked exception declared in your code, you must handle it (in a try/catch block) or declare it (adding 'throws' on method signature).
When you say your method may throw an exception, everyone using your method must handle or declare that exception. If you handle it, you are making your code capable of recovering. If you declare it, you are passing the problem to the caller.
The compiler is fine if you declare 'throws', he knows what to do when your method will be called.
If you declare an exception that is not actually there, you are making users of your code aware that, in a future release, you may add that exception. Users will be prepared for the day you will add that exception.
Specifying says that method can throw an exception or can not throw. But compiler don't check it, because he can't, actually. So try block only check specifying. The
point is compiler don't see deference between actual calling exception and method's specifying. If you call test() in try block, it'll valid. Sorry for my English)
The following code compiles and runs on Java 13:
public class CheckedExceptionSSCE {
public static void main(String[] args) {
try {
methodNoThrowsDeclaration();
} catch (Exception e) {
// why is this throw allowed?
// no throws in main()
throw e;
}
}
private static void methodNoThrowsDeclaration() {
System.out.println("doesn't throw");
}
}
How come the throw e is allowed?
Is it specified somewhere in the JLS? I was not able to find it, perhaps I'm using wrong keywords to search.
Is the compiler smart enough to deduce that there will be no real checked exception thrown and thus allows the code to compile and run?
This is a feature that was added in Java 7.
The compiler can derive the type of exception if you use a variable from the catch clause to rethrow the exception. Since you have no checked exception to be caught, it knows that e could only be RuntimeException and no throws definition is needed.
More information:
https://docs.oracle.com/javase/7/docs/technotes/guides/language/catch-multiple.html
The method:
private static void methodNoThrowsDeclaration() {
System.out.println("doesn't throw");
}
does not throw any checked exception, from this SO thread one can read:
You can throw unchecked exceptions without having to declare them if
you really want to. Unchecked exceptions extend RuntimeException.
therefore there is no need to adapt the main method signature.
From the Java language specification (§14.18 The throw Statement) one can read:
ThrowStatement: throw Expression ;
At least one of the following three conditions must be true, or a
compile-time error occurs:
The type of the Expression is an unchecked exception class (§11.1.1)
or the null type (§4.1).
The throw statement is contained in the try block of a try statement
(§14.20) and it is not the case that the try statement can throw an
exception of the type of the Expression. (In this case we say the
thrown value is caught by the try statement.)
The throw statement is contained in a method or constructor
declaration and the type of the Expression is assignable (§5.2) to at
least one type listed in the throws clause (§8.4.6, §8.8.5) of the
declaration.
The code that you have shown follows at least the first condition. However, let us look at the following example, where non of those three aforementioned conditions are true, namely if you do :
private static void methodNoThrowsDeclaration() {
System.out.println("doesn't throw");
throw new Exception();
}
that would force you to do:
private static void methodNoThrowsDeclaration() throws Exception {
System.out.println("doesn't throw");
throw new Exception();
}
which in turn would give a compiler error:
Unhandled exception: java.lang.Exception
at the statement throw e;.
I have a hard time to understand why the following code compiles, while it is not a subclass of exception:
class Test
{
public void run() throws IOException
{
System.out.println("Test");
}
}
class SubTest extends Test
{
//not a subclass of IOException, still compiles
public void run() throws RuntimeException
{
System.out.println("Test from sub");
}
}
class Sub2Test extends Test
{
//not a subclass of IOException, does not compile
public void run() throws Exception
{
System.out.println("Test from sub");
}
}
I understand RuntimeException is an unchecked exception, but I thought the rule was that it must be a subclass of the parent exception?
Imagine there is a caller which calls Test#run. In the declaration of Test#run, it says it might throw IOException, so the caller knows it can catch and handle it:
Test test = // it could be instance of SubTest of Sub2Test
try {
test.run();
} catch (IOException e) {
}
Then it's ok if SubTest does not throw IOException, the caller will not miss anything.
But if you throw some checked Exception like Sub2Test, since the caller does not know it until runtime, the called is not able to catch and handle it. So it should not be compiled.
"I understand RuntimeException is an unchecked exception, but I thought the rule was that it must be a subclass of the parent exception?"
That is the general rule, as specified in the JLS in section §11.2. Compile-Time Checking of Exceptions, which states (emphasis mine)
The throws clause of an overriding method may not specify that this method will result in throwing any checked exception which the overridden method is not permitted, by its throws clause, to throw (§8.4.8.3).
But that only applies to checked exceptions, and it also explicitly states that
The unchecked exception classes (§11.1.1) are exempted from compile-time checking.
So the compiler is going to ignore the fact that RuntimeException isn't a subclass of IOException.
I have some few doubts on exceptions.
Can anyone tell me why java doesnt allow us to create Checked Exception in a Subclass while it allows Unchecked exception in a subclass
Below exampple throws Compile time error when I use 'throws IOException' , BUT it doesnt throw any error when I use 'throws ArithmeticException' in a subclass.. I just wanna know the actual reason behind it, so can you please?
Here is code (you will get compile time error)
package com.exception.test;
import java.io.IOException;
public class Parent {
void msg() {
System.out.println("Parent...");
}
public static void main(String[] args) {
Parent parent = new Child();
parent.msg();
}
}
class Child extends Parent {
void msg() throws IOException {
System.out.println("Child...");
}
}
//using unCheckedException
package com.exception.test;
import java.io.IOException;
public class Parent {
void msg() {
System.out.println("Parent...");
}
public static void main(String[] args) {
Parent parent = new Child();
parent.msg();
}
}
class Child extends Parent {
void msg() throws ArithmeticException {
System.out.println("Child...");
}
}
If a subclass method declares it can throw a checked exception that the parent doesn't, it breaks the Liskov substitution principle, which is one of the corner stones of object oriented programming.
Consider this bit of code, with Child.msg declared to throw a checked exception:
void doMsg(Parent p) {
p.msg();
}
The program semantics break if you pass in a child object because the checked exception is now neither being caught nor thrown: the exception is no longer "checked."
Since unchecked exceptions can be thrown everywhere, declaring to throw one serves no other purpose than documentation. Therefore it can be allowed safely.
The msg() method in your parent class can throw any unchecked exception it likes. Hence, if you explicitly declare that your child throws an unchecked exception, you're not actually altering the contract. Your child method may throw an ArithmeticException, but so might your parent method.
Checked exception can be narrowed down while overriding, but not can't be broaden. Unchecked exception need not be caught by the overridden methods
From the java specs
The checked exception classes named in the throws clause are part of
the contract between the implementor and user of the method or
constructor. The throws clause of an overriding method may not specify
that this method will result in throwing any checked exception which
the overridden method is not permitted, by its throws clause, to
throw.
Unfortunately, you have hit yet another pitfall of Java's misfeature known as Checked Exceptions. The error you are receiving is an actual problem that all Java professionals face: you are implementing a method with some code that happens to throw a checked exception not declared by the superclass method. The declared checked exceptions are a part of the Java method signature; you can reduce the list in a subclass, but you cannot expand it.
If this is more that just a "why" question, and you need a workaround, the standard idiom is
try {
...code that throws the checked exception...
} catch (TheCheckedException e) { throw new RuntimeException(e); }
This is called exception wrapping. If you have more than one or two undeclared checked exceptions, you can also use the opposite idiom, ensuring that all declared exceptions propagate transparently and all undeclared ones get wrapped:
try {
...code that throws various checked exceptions...
}
catch (DeclaredEx1 | DeclaredEx2 | RuntimeException e) { throw e;}
catch (Exception e) { throw new RuntimeException(e); }
If I have a class called Boat, and if I write :
class Boat throws Exception
Inside my class I am using try catch block to catch ArithmeticException for instance.
What benefit there is to declare an exception versus not declaring an exception?
A class does not throw exceptions. Only methods do.
For some exceptions (checked exceptions that may occur in your code and that you do not catch) the compiler forces you to declare them.
You never have to declare RuntimeExceptions (such as ArithmeticException), but you can. This serves as documentation.
You can declare checked exceptions that your code does not throw. This makes it future-proof if you might later want to throw them, and also allows for subclasses to do such.
When declaring exceptions, you can go broad/generic (throws Exception or even throws Throwable), but it is generally better to be more specific. That gives the people using your code a better idea of what to expect. The whole purpose of having these many specific Exception classes is to make it easier to handle exceptions appropriately (and have the compiler enforce the fact that someone at least thought about doing that).
First classes won't throw Exception, only methods will throw
check this Example
class A
{
public A() throws Exception
{
int k=5/0;
}
}
public class B
{
public static void main(String[] args)
{
A a=new A();
}
}
When you run the above you'll get compile time error saying
Unhandled exception type Exception
because when ever methods throws exception , they are expected to handle by calling object, it is just like saying 'Hey somebody handle Exception', so the method calling it should handle the Exception , if they don't it will be a compile time error.
For the smooth execution of program, Exception thrown by method should be handle by calling method. Here is how you need to handle.
class A
{
public A() throws Exception
{
int k=5/0;
}
}
public class B
{
public static void main(String[] args)
{
try{
A a=new A();
}
catch(Exception e)
{
System.out.println("caught "+e);
}
}
}
Declaring Exception will allow your program to compile if any Exception can be determined by the compiler to "escape". It's primary purpose, though, is to document to the user of your method (and to their compilers) that it may throw an Exception.
As such, declaring Exception, vs declaring a more specific exception cause, is nullifying the benefits of the scheme -- cheating.
If you write something like that,
double x = 1/0;
System.out.print("hey");
it will give you an exception and your code is going to stop, will not print hey. But if you encapsulate it with try/catch, it will not stop.
try{
double x = 1/0;
}catch(ArithmeticException){}
System.out.print("hey");
This will print hey.