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more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed share|improve this answer answered Jan 25 '14 at 14:01 Sergiu 695 add a comment| up vote 0 down vote Here's an example public static void main(String[] args) { System.out.println(add5(1)); } public JVM Troubleshooting Guide3. Join them; it only takes a minute: Sign up What actually causes a Stack Overflow error? [duplicate] up vote 205 down vote favorite 39 This question already has an answer here: http://cpresourcesllc.com/stack-overflow/stack-overflow-error-java.php

To expand on an example that utilizes an object with methods we can look here: if(dog == null || dog.firstName == null) return; The above is the proper order to check How to deal with the StackOverflowErrorThe simplest solution is to carefully inspect the stack trace and detect the repeating pattern of line numbers. Depending on the Java Virtual Machine (JVM) installed, the default thread stack size may equal to either 512KB, or 1MB. It may reserve some stack space for temporary storage, but this is often less than the maximum amount of bytecode interpretation stack required. –Jules Mar 5 '14 at 8:09 2 http://stackoverflow.com/questions/22182669/what-actually-causes-a-stack-overflow-error

Stack Overflow Error Java

Update: Bad luck, this is recursion, I asked question about that here. So, nothing will be overwritten. A StackOverflowError, however is thrown when a thread requires an additional frame when there is no room for it. (This is confirmed by comments and answers on this page). –11684 Mar Such reason for 'StackOverflowError' will require much time to rectify it.

A new frame is created and added (pushed) to the top of stack for every method invocation. For example, your app may be handling paint messages and whilst processing them it may call a function that causes the system to send another paint message. Parameters:s - the detail message. Fix Stack Overflow Error Just keep calling methods, without ever returning.

Dec 28 '15 at 21:24 This question has been asked before and already has an answer. As you allocate memory this heap can grow towards the upper end of your address space. How to change 'Welcome Page' on the basis of logged in user or group? http://stackoverflow.com/questions/22182669/what-actually-causes-a-stack-overflow-error Thanks. –JB Nizet Mar 4 '14 at 20:57 2 You got your answer in all the other answers.

Setting break points in your application allows for the step-by-step processing of the stack. Stackoverflowerror Java Recursion Simple Example With the example given in the question, we can determine exactly where the exception was thrown in the application. Not the answer you're looking for? For instance, in the mentioned recursion.

How To Solve Stack Overflow Error In Java

How to deal with Stacktraces/Exceptions? https://docs.oracle.com/javase/7/docs/api/java/lang/StackOverflowError.html As for the corruption of thing, the execution is stopped at a very precise point, just like with any other error. Stack Overflow Error Java Let's use a small example to show why not to just catch all Exceptions: int mult(Integer a,Integer b) { try { int result = a/b return result; } catch (Exception e) How To Resolve Stack Overflow Error In Java your favorite Add new comment Your name Email The content of this field is kept private and will not be shown publicly.

According to JVM specification link (given by JB Nizet in a comment) JVM should throw a StackOverflowError, not die: If the computation in a thread requires a larger Java Virtual Machine http://cpresourcesllc.com/stack-overflow/stack-overflow-error-recursion-java.php I don't follow what they are saying. –retrohacker Mar 4 '14 at 21:19 There are no realistic circumstances where you can assume you are "safe" after catching a StackOverflowError. To elaborate on the second question: When Java throws the StackOverflowError, can you safely assume that the stack did not write into the heap? So recursion is always responsible for stack overflows? Java Increase Stack Size

  1. If a thread requires a new frame and there isn’t enough memory to allocate it then an OutOfMemoryError is thrown.
  2. If there is no memory left in stack for storing function call or local variable, JVM will throw java.lang.StackOverFlowError while if there is no more heap space for creating object, JVM
  3. This works because tail-recursion calls do not take up additional stack space.[3] C compiler options will effectively enable tail-call optimization; compiling the above simple program using gcc with -O1 will result
  4. try { Socket x = new Socket("1.1.1.1", 6789); x.getInputStream().read() } catch (IOException e) { System.err.println("Connection could not be established, please try again later!") } Why should I not use catch (Exception
  5. Lagrange multiplier on unit sphere Enigmatic Movie Riddle ¿Cuál es la razón por la que se corrije "yo y tú" a "tú y yo"?

See docs.oracle.com/javase/specs/jvms/se7/html/… –JB Nizet Mar 4 '14 at 22:35 1 @Ingo the bytecode stack is somewhat distinct from the stack used at runtime for method calls. I don't follow what they are saying. –retrohacker Mar 4 '14 at 21:19 There are no realistic circumstances where you can assume you are "safe" after catching a StackOverflowError. It cannot corrupt anything, in fact it is very possible to catch it and recover from it. http://cpresourcesllc.com/stack-overflow/stack-overflow-error-in-java.php The bottom-most "Caused by" will be the root.

Also see the documentation redistribution policy. Stack Overflow In Java With Example It may have done only part of an operation. I got something unexpected: # # A fatal error has been detected by the Java Runtime Environment: # # SIGSEGV (0xb) at pc=0x00007f4822f9d501, pid=4988, tid=139947823249152 # # JRE version: 6.0_27-b27 #

I'll see what I can do. –Rob Hruska Oct 21 '10 at 16:26 1 @Rob Hruska, sorry, I meant that you can potentially have more than 1 "Caused By" in

As for the corruption of thing, the execution is stopped at a very precise point, just like with any other error. Should I report this? If you can't do that though, the second best thing would be to look whether there's something that clearly causes the stack overflow. Stackoverflow Java Tutorial It may reserve some stack space for temporary storage, but this is often less than the maximum amount of bytecode interpretation stack required. –Jules Mar 5 '14 at 8:09 2

In the catch part, we've used the Exception type that was reported in the stack trace - ArithmeticException. I got something unexpected: # # A fatal error has been detected by the Java Runtime Environment: # # SIGSEGV (0xb) at pc=0x00007f4822f9d501, pid=4988, tid=139947823249152 # # JRE version: 6.0_27-b27 # Would you like to answer one of these unanswered questions instead? Check This Out asked 2 years ago viewed 46967 times active 1 year ago Linked 233 What is a StackOverflowError? 0 Stack Overflow error in TicTacToe Java 21 What's the difference between StackOverflowError and

Career OpportunitiesKnowledge BaseCoursesNewsResourcesTutorialsWhitepapersThe Code Geeks Network.NET Code GeeksJava Code GeeksSystem Code GeeksWeb Code GeeksHall Of FameAndroid Alert Dialog ExampleAndroid OnClickListener ExampleHow to convert Character to String and a String to Character functors[0].fun(); // Sorry, couldn't resist to not comment in such moment. } interface Functor { void fun(); } } Compile with standard javac Test.java and run with java -Xss104k Test 2> We want to create new Test object... Just like the reason we call it 'stack' is because stack is First in Last out (FILO), the deepest exception was happened in the very beginning, then a chain of exception

SO in not an AOOBE, but it can be compared to one because basically you are trying to add an element to a stack that is full. Hotspot doesn't do it, at least by default. Once you detect these lines, you must carefully inspect your code and understand why the recursion never terminates. There isn't a fixed number of method calls, there isn't anything specific about recursion, you get the exception which you try to call a method and there isn't enough memory.

The call stack may consist of a limited amount of address space, often determined at the start of the program. Oh man: bad times. –Ziggy Oct 18 '08 at 9:49 add a comment| up vote 4 down vote The most common cause of stack overflows is excessively deep or infinite recursion. A method with five local variables and three parameters will use more stack space than a void doStuff() method with no local variables will. Most compiled languages have tail-recursion optimizations.

A stacktrace is a very helpful debugging tool. So, in case of OpenJDK 20.0-b12 we can see that JVM firstly exploded. The stack memory is used to store local variables and function call, while heap memory is used to store objects in Java. So the currently executing method is at the top of the stack.

Add a language to a polyglot Secret salts; why do they slow down attacker more than they do me? It's a very specific error so no. But what it also catches is a possible NullPointerException that is thrown if a or b are null. Email address: Join Us With 1,240,600 monthly unique visitors and over 500 authors we are placed among the top Java related sites around.

If you call some recursive function, and it throws back a StackOverflowError you can make no assumptions about the correctness of the state of any objects that it touched anymore. In itself it will not write stuff in the heap, which is the sense of the question (can you safely assume that the stack did not write into the heap?) –njzk2