What does this error mean?
Note: Main.java uses unchecked or unsafe operations.
Note: Recompile with -Xlint:unchecked for details.
Any suggestions how to avoid that kind of error?
That's the error you get when you use Collections without specifying a type. You probably have something like:
ArrayList myList = new ArrayList(); // or some other Collection class
If that's the case, you need to change that to specify what type of objects you want to store. For example:
ArrayList<String> myList = new ArrayList<String>();
Read up on Java Generics for more information.
This is my best guess without seeing your code and the full error message. There could be other causes for that message, this is just the problem that I've seen accompany that message before.
First, recompile with -Xlint:unchecked to see what the problem is. Then fix those problems. There are a number of potential high-level causes for unchecked warnings. One is that you didn't provide type parameters where you should have. There are some situations where they are unavoidable, and then you can suppress the specific warning, but these are the exception, and care must be taken that you aren't suppressing warnings that are really important.
So recompile with -Xlint:unchecked and post additional questions if you have trouble with any of the specific issues that are revealed.
What does it mean?
Java generics allow you to write something like this:
List<String> l = ...;
String s = l.get(0); // note there is no explicit typecast.
But if the compiler tells you that your code has "unchecked or unsafe operations", it is saying that you have broken the rules for using generics safely, and your code may give runtime class cast exceptions at unexpected places; e.g. in the statement above where we left out the typecast.
There are a few things that will cause the compiler to complain about unchecked or unsafe operations, and each one requires a different remediation. Do what the compiler is telling you and run it with the -Xlint option.
Related
Below is an excerpt from java documentation here
Because of type erasure, List<Number> and List<String> both become List. Consequently, the compiler allows the assignment of the object l, which has a raw type of List, to the object ls.
Also from the same documentation
Consider the following example:
List l = new ArrayList<Number>();
List<String> ls = l; // unchecked warning
l.add(0, new Integer(42)); // another unchecked warning
String s = ls.get(0); // ClassCastException is thrown
During type erasure, the types ArrayList and List become ArrayList and List, respectively.
Why compiler can't show this as an error (instead of warning). Could someone kindly help with an example where warning is needed and will cause an issue (or make certain useful patterns impossible) when warning becomes an error.
Java gained type variables in java 1.5. Before java 1.5, they did not exist.
Java also strongly dislikes releasing new versions such that 'upgrading' is more involved than just 'simply compile your existing code with the new compiler and it all just works exactly like it did before'. Also, 'you can run code compiled with the javac from java 1.4, on the java.exe from 1.5 and it runs fine'.
Why? Well, because otherwise you get a split community, and because most projects use 500 dependencies, that is hugely detrimental. You can't 'upgrade' from java 1.4 to 1.5 until each and every single last one of those 500 deps has upgraded.
Combine these 2 facts and generics makes sense:
Why erasure at all? Because how else could it work? Class files stemming from javac1.4 cannot possibly have the generics info, given that they did not exist when javac1.4 was released!
Why 'raw types'? Because pre-generics code by their nature have them (List<T> would be a compiler error on java 1.4), and by their nature, there is no sanity, typevar-wise. You could write this:
List foo = new ArrayList();
foo.add(5);
foo.add("hello");
and it compiles just fine on javac 1.4. Therefore it MUST also compile on javac 1.5. As a compromise, if you attempt to compile that on javac from 1.5, it works and produces a class file that acts identically, but you do get warnings.
java 1.5 is now 25 years old or whatnot. So it all seems weird and archaic now. But you're asking 'why'. This is why.
The following code compiles and runs successfully without any exception
import java.util.ArrayList;
class SuperSample {}
class Sample extends SuperSample {
#SuppressWarnings("unchecked")
public static void main(String[] args) {
try {
ArrayList<Sample> sList = new ArrayList<Sample>();
Object o = sList;
ArrayList<SuperSample> ssList = (ArrayList<SuperSample>)o;
ssList.add(new SuperSample());
} catch (Exception e) {
e.printStackTrace();
}
}
}
shouldn't the line ArrayList<SuperSample> ssList = (ArrayList<SuperSample>)o; produce a ClassCastException ?
while the following code produces a compile time error error to prevent heap pollution, shouldn't the code mentioned above hold a similar prevention at runtime?
ArrayList<Sample> sList = new ArrayList<Sample>();
ArrayList<SuperSample> ssList = (ArrayList<SuperSample>) sList;
EDIT:
If Type Erasure is the reason behind this, shouldn't there be additional mechanisms to prevent an invalid object from being added to the List? for instance
String[] iArray = new String[5];
Object[] iObject = iArray;
iObject[0]= 5.5; // throws ArrayStoreException
then why,
ssList.add(new SuperSample());
is not made to throw any Exception?
No it should not, at run time both lists have the same type ArrayList. This is called erasure. Generic parameters are not part of compiled class, they all are erased during compilation. From JVM's perspective your code is equal to:
public static void main(String[] args) {
try {
ArrayList sList = new ArrayList();
Object o = sList;
ArrayList ssList = (ArrayList)o;
ssList.add(new SuperSample());
} catch (Exception e) {
e.printStackTrace();
}
}
Basically generics only simplify development, by producing compile time errors and warnings, but they don't affect execution at all.
EDIT:
Well, the base concept behind this is Reifiable Type. Id strongly recomend reading this manual:
A reifiable type is a type whose type information is fully available
at runtime. This includes primitives, non-generic types, raw types,
and invocations of unbound wildcards.
Non-reifiable types are types where information has been removed at
compile-time by type erasure
To be short: arrays are rifiable and generic collections are not. So when you store smth in the array, type is checked by JVM, because array's type is present at runtime. Array represents just a piece of memmory, while collection is an ordinary class, which might have any sort of implementation. For example it can store data in db or on the disk under the hood. If you'd like to get deeper, I suggest reading Java Generics and Collections book.
In your code example,
class SuperSample { }
class Sample extends SuperSample { }
...
ArrayList<Sample> sList = new ArrayList<Sample>();
Object o = sList;
ArrayList<SuperSample> ssList = (ArrayList<SuperSample>)o;
Shouldn't the last line produce a ClassCastException?
No. That exception is thrown by the JVM when it detects incompatible types being cast at runtime. As others have noted, this is because of erasure of generic types. That is, generic types are known only to the compiler. At the JVM level, the variables are all of type ArrayList (the generics having been erased) so there is no ClassCastException at runtime.
As an aside, instead of assigning to an intermediate local variable of type Object, a more concise way to do this assignment is to cast through raw:
ArrayList<SuperSample> ssList = (ArrayList)sList;
where a "raw" type is the erased version of a generic type.
Shouldn't there be additional mechanisms to prevent an invalid object from being added to the List?
Yes, there are. The first mechanism is compile-time checking. In your own answer you found the right location in the Java Language Specification where it describes heap pollution which is the term for an invalid object occurring in the list. The money quote from that section, way down at the bottom, is
If no operation that requires a compile-time unchecked warning to be issued takes place, and no unsafe aliasing occurs of array variables with non-reifiable element types, then heap pollution cannot occur.
So the mechanism you're looking for is in the compiler, and the compiler notifies you of this via compilation warnings. However, you've disabled this mechanism by using the #SuppressWarnings annotation. If you were to remove this annotation, you'd get a compiler warning at the offending line. If you absolutely want to prevent heap pollution, don't use #SuppressWarnings, and add the options -Xlint:unchecked -Werror to your javac command line.
The second mechanism is runtime checking, which requires use of one of the checked wrappers. Replace the initialization of sList with the following:
List<Sample> sList = Collections.checkedList(new ArrayList<Sample>(), Sample.class);
This will cause a ClassCastException to be thrown at the point where a SuperSample is added to the list.
The key here to answer your question is Type Erasure in java
You have a warning at compile time for your first case and not in the second because of your indirection by an object which prevent the compiler to raise you a warning (I'm guessing that this warning is raised when casting a parametrized type to another one which is not done on your second case, if anyone can confirm that I would be glad to here about it).
And your code run because, in the end sList ssList et o are all ArrayList
I think that this cant produce ClassCastException because of backward compatibility issue in Java.
Generic information is not included in bytecode (compiler get rids of it during compilation).
Imagine scenario that you use in your project some old legacy code (some old library writen in java 1.4) and you pass generic List to some method in this legacy code.
You can do this.
In time before generics legacy code was allowed to put anything at all (except primitives) into a collection.
So this legacy code cant get ClassCastException even if it try to put String to List<Integer>.
From the legacy code perspective it is just List.
So this strange behaviour is a consequence of type erasure and to allow backward compatibility in Java.
EDIT:
You get ArrayStoreException for arrays because at runtime the JVM KNOWS the type of arrays, and you dont get any exception for collections because of type erasure and this backward compatibility issue JVM doesnt know the type of collection at runtime.
You can read about this topic in "SCJP Sun® Certified Programmer for Java™ 6 Study Guide" book in chapter 7 "Generics and Collections"
From the JLS (4.12.2)
It is possible that a variable of a parameterized type refers to an object that is not
of that parameterized type. This situation is known as heap pollution. This situation
can only occur if the program performed some operation that would give rise
to an unchecked warning at compile-time.
For example, the code:
List l = new ArrayList<Number>();
List<String> ls = l; // unchecked warning
gives rise to an unchecked warning, because it is not possible to ascertain, either at compile-
time (within the limits of the compile-time type checking rules) or at run-time, whether
the variable l does indeed refer to a List<String>.
If the code above is executed, heap pollution arises, as the variable ls, declared to be a
List<String>, refers to a value that is not in fact a List<String>.
The problem cannot be identified at run-time because type variables are not reified,
and thus instances do not carry any information at run-time regarding the actual type
parameters used to create them.
This question already has answers here:
What is SuppressWarnings ("unchecked") in Java?
(11 answers)
Closed 9 years ago.
I am learning Java. I have task to implement HashDictionary using DictionaryInterface
Hashtable<K,V> h;
Set<K> s = h.keySet();
K elements[]=(K[])Array.newInstance(KeyType, h.size());
When I have written the above statements the Eclipse IDE is showing a warning message.
#SuppressWarnings("unchecked")
If I add the above message before start of a method. The warning is disappeared. What is this means. Can anyone give the reason? Thanks in Advance
Warning in general are just that - warnings. Things that can easily go wrong, but havent technically gone wrong yet. Sometimes you are well aware of them and can simply ignore them. In those cases you might want to use #SuppressWarnings("unchecked"). That will signal the compiler to no post its warning, thus giving you an absolutely clean compile.
Note that even if you dont put that there and you do get the warning your program can still work just fine. Its just that there is an elevated chance that your program will break at that point.
If you want, you can also remove the warning entirely from eclipse by going to Windows > Preferences > General > Editors > Text Editors > Annotations > Warnings > Annotation types and then select the warnings you do/dont want.
The annotation type SuppressWarnings supports programmer control over warnings otherwise issued by the Java compiler. It contains a single element that is an array of String. If a program declaration is annotated with the annotation #SuppressWarnings(value = {someString}), then a Java compiler must not report any warning identified by one of someString if that warning would
have been generated as a result of the annotated declaration or any of its parts.
For example, Unchecked warnings are identified by the string unchecked. In your example, the below statement will emit an unchecked warning:
K elements[]=(K[])Array.newInstance(KeyType, h.size());
This statement contains an unchecked conversion, as the compiler doesn't know if the cast is safe or not; and will simply emit an unchecked conversion warning. Which can be suppressed by using #SuppressWarnings("unchecked") annotation.
In case you don't already know (you said you're a beginner), Eclipse often tells you what is causing the warning and can sometimes give you suggestions on how to fix it.
Of course, it's always much better to fix the warnings than to suppress them (unless you're doing really intricate stuff). Hope this helps.
I have below code snippet and this works fine. Shouldn't it throw compile time error because I have defined c as ArrayList which will contain String object but I am adding Integer object. So why it did not throw compile time/Run time error?
Collection c = new ArrayList<String>();
c.add(123);
I know below will throw compile time error but why not above. Whats the logical difference between both these code snippet?
Collection<String>() c = new ArrayList();
c.add(123);
The first code snippet does not result in a compile time error, because at the line
c.add(123)
the compiler inspects the type of c. Since you declared c as Collection, the compiler treats it as such. Since Collection offers a method add(Object), it is perfectly reasonable to add any object to the c, especially an integer. Note that this program will however result in a runtime-error, if you attempt to read back the collection values as Strings.
In your second code snippet you provide more information for the compiler to work with. In this snippet it knows that the Collection it deals with is an Collection<String>, which can only accept Strings. Thus, there is no method add(int) or add(Object), only add(String). This leads to a compile-time error.
why it did not throw compile time error?
Because it's not syntactically or semantically invalid, it's just unwise.
Note that most modern IDEs (e.g. Eclipse) can be configured to warn you about the unparameterised Collection c, and optionally to fail to compile.
In the first example, the collection is "raw". This will usually result in a warning but not an error (depending on your exact set-up). This is primary in order to be able to compile all the pre-Java 5 legacy code around.
The the second example, you assign a "raw" object to a parameterized version, which only can be done with an explicit cast.
1) What is the logical difference?
Above: A Collection can be declared without a generic type. This is called a raw type. The collection can then hold any kind of collection. Since, with a raw typed collection, at runtime you might use a collection of strings as a collection of integers causing a runtime exception the compiler will usually throw a warning. Since you have not typed the collection in the above example the compiler can not prevent these runtime exceptions. The warning can be ignored if you know what it is for and know what you are doing.
Below: But a variable declared as a Collection<String> cannot hold any kind of collection. It has to be a collection of the type String. It is strong typed. The compiler is correct to see this as an error.
2) Why does the above snippet not cause a compiler error?
Java is strong typed, which ensures type safety. The above snippet is not type safe, but allowed by Java nonetheless. This is probably for historical reasons: Generics were only introduced with Java 1.5, so if the above snippet would have caused a compile error then most Java 1.4 code would have been broken in the Java 1.5 compiler.
Not every programming language evolves in such a backward compatible manner (PHP for instance). Apparently backward compatibility was valued over type safety when introducing Java 1.5.
I have the following code
String innerText = null;
innerText = this.getException(detail.getChildElements());
causing this warning
Type safety: The expression of type Iterator needs unchecked conversion to conform
to Iterator
The referenced method is
private String getException(Iterator<OMElementImpl> iterator) { ... }
The other method, getChildElements(), is in a JAR file that I can't touch. There are no other warnings or errors.
From Googling, it seems like the usual way to get rid of this sort of warning is
#SuppressWarnings("unchecked")
String innerText = this.getException(detail.getChildElements());
because the compiler can't guarantee safety ahead of time, but I'd prefer to avoid using SuppressWarnings if possible... is there a better way?
EDIT: getChildElements() is documented here
You can suppress the warning, but if you do so, you are relying 100% on the third-party library, and discarding the assurance of Java generic types: that any ClassCastException raised at runtime will occur right at an explicit cast.
Our coding standard is to suppress warnings only when we can prove the code is type safe—and we treat any calls outside the package as a black box, and don't rely on any comments about the content of a raw collection. So, suppression is extremely rare. Usually, if the code is type safe, the compiler can determine it, although sometimes we have to give it some help. The few exceptions involve arrays of generic type that don't "escape" from a private context.
If you don't fully trust the third-party library, create a new collection, and add the contents after casting them to OMEElementImpl. That way, if there is a bug in the library, you find out about it right away, rather than having some code far distant in time and space blow up with a ClassCastException.
For example:
Iterator<?> tmp = detail.getChildElements();
Collection<OMElementImpl> elements = new ArrayList<OMElementImpl>();
while (tmp.hasNext())
elements.add((OMElementImpl) tmp.next()); /* Any type errors found here! */
String innerText = getException(elements.iterator());
Remember, generics were not invented to make code look pretty and require less typing! The promise of generics is this: Your code is guaranteed to be type-safe if it compiles without warnings. That is it. When warnings are ignored or suppressed, code without a cast operator can mysteriously raise a ClassCastException.
Update: In this case, especially, it seems extremely risky to assume that the result of getChildElements is a iterator of OMElementImpl. At best, you might assume that they are OMElement, and that's only implied from the class, not anything on the method in particular.