How to handle StackOverflowError in Java?

问题:

How to handle StackOverflowError in Java ?

回答1:

I'm not sure what you mean with "handle".

You can certainly catch that error:

public class Example {
    public static void endless() {
        endless();
    }

    public static void main(String args[]) {
        try {
            endless();
        } catch(StackOverflowError t) {
            // more general: catch(Error t)
            // anything: catch(Throwable t)
            System.out.println("Caught "+t);
            t.printStackTrace();
        }
        System.out.println("After the error...");
    }
}

but that is most likely a bad idea, unless you know exactly what you are doing.

回答2:

You probably have some infinite recursion going on.

I.e. a method that calls itself over and over

public void sillyMethod()
{
    sillyMethod();
}

One to handle this is to fix your code so that the recursion terminates instead of continuing forever.

回答3:

Take a look at Raymond Chen's post When debugging a stack overflow, you want to focus on the repeating recursive part. An extract:

If you go hunting through your defect tracking database trying to see whether this is a known issue or not, a search for the top functions on the stack is unlikely to find anything interesting. That's because stack overflows tend to happen at a random point in the recursion; each stack overflow looks superficially different from every other one even if they are the same stack overflow.

Suppose you're singing the song Frère Jacques, except that you sing each verse a few tones higher than the previous one. Eventually, you will reach the top of your singing range, and precisely where that happens depends on where your vocal limit lines up against the melody. In the melody, the first three notes are each a new "record high" (i.e., the notes are higher than any other note sung so far), and new record highs appear in the three notes of the third measure, and a final record high in the second note of the fifth measure.

If the melody represented a program's stack usage, a stack overflow could possibly occur at any of those five locations in the program's execution. In other words, the same underlying runaway recursion (musically represented by an ever-higher rendition of the melody) can manifest itself in five different ways. The "recursion" in this analogy was rather quick, just eight bars before the loop repeated. In real life, the loop can be quite long, leading to dozens of potential points where the stack overflow can manifest itself.

If you are faced with a stack overflow, then, you want to ignore the top of the stack, since that's just focusing on the specific note that exceeded your vocal range. You really want to find the entire melody, since that's what's common to all the stack overflows with the same root cause.

回答4:

You might want to see if the "-Xss" option is supported by your JVM. If so, you might want to try setting it to a value of 512k (default is 256k under 32-bit Windows and Unix) and see if that does anything (other than make you sit longer until your StackOverflowException). Note that this is a per-thread setting, so if you've got a lot of threads running you also might want to bump up your heap settings.

回答5:

The correct answer is the one already given. You likely either a) have a bug in your code leading to an infinite recursion which is usually quite easy to diagnose and fix, or b) have code which can lead to very deep recursions for example recursively traversing an unbalanced binary tree. In the latter situation, you need to alter your code to not allocate the information on the stack (i.e. to not recurse) but to instead allocate it in the heap.

For example, for an unbalanced tree traversal, you could store the nodes that will need to be revisited in a Stack data structure. For an in order traversal you would loop down the left branches pushing each node as you visited it until you hit a leaf, which you would process, then pop a node off the top of the stack, process it, then restart your loop with the right child (by just setting your loop variable to the right node.) This will use a constant amount of stack by moving everything that was on the stack to the heap in the Stack data structure. Heap is typically much more plentiful than stack.

As something that is usually an extremely bad idea, but is necessary in cases where memory use is extremely constrained, you can use pointer reversal. In this technique, you encode the stack into the structure you are traversing, and by reusing the links you are traversing, you can do this with no or significantly less additional memory. Using the above example, instead of pushing nodes when we loop, we just need to remember our immediate parent, and at each iteration, we set the link we traversed to the current parent and then the current parent to the node we are leaving. When we get to a leaf, we process it, then go to our parent and then we have a conundrum. We don't know whether to correct the left branch, process this node, and continue with the right branch, or to correct the right branch and go to our parent. So we need to allocate an extra bit of information as we iterate. Typically, for low-level realizations of this technique, that bit will be stored in the pointer itself leading to no additional memory and constant memory overall. This is not an option in Java, but it may be possible to squirrel away this bit in fields used for other things. In the worst-case, this is still at least 32 or 64 times reduction in the amount of memory needed. Of course, this algorithm is extremely easy to get wrong with completely confusing results and would raise utter havoc with concurrency. So it's almost never worth the maintenance nightmare except where allocating memory is untenable. The typical example being a garbage collector where algorithms like this are common.

What I really wanted to talk about, though, is when you might want to handle the StackOverflowError. Namely to provide tail call elimination on the JVM. One approach is to use trampoline style where instead of performing a tail call you return a nullary procedure object, or if you are just returning a value you return that. [Note: this requires some means of saying a function returns either A or B. In Java, probably the lightest way to do this is to return one type normally and throw the other as an exception.] Then whenever you call a method, you need to do a while loop calling the nullary procedures (which will themselves return either a nullary procedure or a value) until you get a value. An endless loop will become a while loop that is constantly forcing procedure objects that return procedure objects. The benefits of trampoline style is that it only uses a constant factor more stack than you would use with an implementation that properly eliminated all tail calls, it uses the normal Java stack for non-tail calls, the translation simple, and it only grows the code by a (tedious) constant factor. The drawback is you allocate an object on every method call (which will immediately become garbage) and consuming these objects involves a couple of indirect calls per tail call.

The ideal thing to do would be to never allocate those nullary procedures or anything else in the first place, which is exactly what tail call elimination would accomplish. Working with what Java provides though, what we could do is run the code as normal and only make these nullary procedures when we run out of stack. Now we still allocate those useless frames, but we do so on the stack rather than the heap and deallocate them in bulk, also, our calls are normal direct Java calls. The easiest way to describe this transformation is to first rewrite all multi-call-statement methods into methods that have two call statements, i.e. fgh() { f(); g(); h(); } becomes fgh() { f(); gh(); } and gh(){ g(); h(); }. For simplicity, I'll assume all methods end in a tail call, which can be arranged by just packaging the remainder of a method into a separate method, though in practice, you'd want to handle these directly. After these transformations we have three cases, either a method has zero calls in which case there is nothing to do, or it has one (tail) call, in which case we wrap it in a try-catch block in the same we will for the tail call in the two call case. Finally, it may have two calls, a non-tail call and a tail call, in which case we apply the following transformation illustrated by example (using C#'s lambda notation which could easily be replaced with an anonymous inner class with some growth):

// top-level handler
Action tlh(Action act) {
    return () => {
        while(true) {
            try { act(); break; } catch(Bounce e) { tlh(() => e.run())(); }
        }
    }
}

gh() {
    try { g(); } catch(Bounce e) { 
        throw new Bounce(tlh(() => { 
            e.run(); 
            try { h(); } catch(StackOverflowError e) {
                throw new Bounce(tlh(() => h());
            }
        }); 
    }
    try { h(); } catch(StackOverflowError e) { 
        throw new Bounce(tlh(() => h())); 
    }
}

The main benefit here is if no exception is thrown, this is the same code as we started with just with some extra exception handlers installed. Since tail calls (the h() call) don't handle the Bounce exception, that exception will fly through them unwinding those (unnecessary) frames from the stack. The non-tail calls catch the Bounce exceptions and rethrow them with the remaining code added. This will unwind the stack all the way up to the top level, eliminating the tail call frames but remembering the non-tail call frames in the nullary procedure. When we finally execute the procedure in the Bounce exception at the top-level, we will recreate all the non-tail call frames. At this point, if we immediately run out of stack again, then, since we don't reinstall the StackOverflowError handlers, it will go uncaught as desired, since we really are out of stack. If we get a little further, a new StackOverflowError will be installed as appropriate. Furthermore, if we do make progress, but then do run out of stack again, there is no benefit re-unwinding the frames we already unwound, so we install new top-level handlers so that the stack will only be unwound up to them.

The biggest problem with this approach is that you'll probably want to call normal Java methods and you may have arbitrarily little stack space when you do, so they may have enough space to start but not finish and you can't resume them in the middle. There are at least two solutions to this. The first is to ship all such work to a separate thread which will have it's own stack. This is pretty effective and pretty easy and won't introduce any concurrency (unless you want it to.) Another option is simply to purposely unwind the stack before calling any normal Java method by simply throwing a StackOverflowError immediately before them. If it still runs out of stack space when you resume, then you were screwed to begin with.

A similar thing can be done to make continuations just-in-time too. Unfortunately, this transformation isn't really bearable to do by hand in Java, and is probably borderline for languages like C# or Scala. So, transformations like this tend to be done by languages that target the JVM and not by people.

回答6:

I guess you can't - or it at least depends on the jvm you use. Stack overflow means, that you have no room to store local variables and return adresses. If your jvm does some form of compiling, you have the stackoverflow in the jvm as well and that means, you can't handle it or catch it. The jvm has to terminate.

There could be a way to create a jvm that allows for such behavior, but it would be slow.

I have not tested the behavior with the jvm, but in .net you just can't handle the stackoverflow. Even try catch won't help. Since java and .net rely on the same concept (virtual machines with jit) I suspect java would behave the same. The presence of a stackoverflow-exception in .NET suggests, there might be some vm that does enable the program to catch it, the normal does not though.

回答7:

Most chances to get StackOverflowError are by using [long/infinite] recursions in a recursive functions.

You can avoid Function recursion by changing your application design to use stackable data objects. There are coding patterns to convert recursive codes to iterative code blocks. Have a look at below answeres:

• way-to-go-from-recursion-to-iteration
• can-every-recursion-be-converted-into-iteration
• design-patterns-for-converting-recursive-algorithms-to-iterative-ones

So, you avoid memory stacking by Java for your recessive function calls, by using your own data stacks.

回答8:

The stack trace should indicate the nature of the problem. There should be some obvious looping when you read the stack trace.

If it's not a bug, you need add a counter or some other mechanism to halt the recursion before the recursion goes so deep it causes a stack overflow.

An example of this might be if you're handling nested XML in a DOM model with recursive calls and the XML is nested so deep it causes a stack overflow with your nested calls (unlikely, but possible). This would have to be pretty deep nesting to cause a stack overflow though.

回答9:

As mentioned by many in this thread, the common cause for this is a recursive method call that doesn't terminate. Where possible avoid the stack overflow and if you this in testing you should consider this in most cases to be a serious bug. In some cases you can configure the thread stack size in Java to be larger to handle some circumstances ( large data sets being managed in local stack storage, long recursive calls) but this will increase the overall memory footprint which can lead to issues in the number of threads available in the VM. Generally if you get this exception the thread and any local data to this thread should be considered toast and not used( ie suspect and possibly corrupt).

回答10:

Simple,

Look at the stack trace that the StackOverflowError produces so you know where in your code it occurs and use it to figure out how to rewrite your code so that it doesn't call itself recursively (the likely cause of your error) so it won't happen again.

StackOverflowErrors are not something that needs to be handled via a try...catch clause but it points to a basic flaw in the logic of your code that needs to be fixed by you.

回答11:

java.lang.Error javadoc:

An Error is a subclass of Throwable that indicates serious problems that a reasonable application should not try to catch. Most such errors are abnormal conditions. The ThreadDeath error, though a "normal" condition, is also a subclass of Error because most applications should not try to catch it. A method is not required to declare in its throws clause any subclasses of Error that might be thrown during the execution of the method but not caught, since these errors are abnormal conditions that should never occur.

So, don't. Try to find what's wrong in the logic of your code. This exception ocurrs very often because of infinite recursion.

回答12:

in some occasion, you can't catch stackoverflowerror. whenever you try, you will encounter new one. because is java vm. it's good to find recursive code blocks, like Andrew Bullock's said.

回答13:

/*
Using Throwable we can trap any know error in JAVA..
*/
public class TestRecur {
    private int i = 0;


    public static void main(String[] args) {
        try {
            new TestRecur().show();
        } catch (Throwable err) {
            System.err.println("Error...");
        }
    }

    private void show() {
        System.out.println("I = " + i++);
        show();
    }
}

However you may have a look at the link: http://marxsoftware.blogspot.in/2009/07/diagnosing-and-resolving.html to understand the code snippet, which may raise error