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Java.lang.stackoverflowerror Android

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A simple example is public static void main(String... If there is no space for a new stack frame then, the StackOverflowError is thrown by the Java Virtual Machine (JVM). share|improve this answer answered Oct 18 '08 at 8:19 Greg 199k35307303 Oops, didn't see the Java tag –Greg Oct 18 '08 at 8:23 Also, from the original If you've got no obvious recursive functions then check to see if you're calling any library functions that indirectly will cause your function to be called (like the implicit case above). Source

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Java.lang.stackoverflowerror Android

BrandPostsLearn more Sponsored by IBM Tooling the Optimal Hybrid Landscape A recent What is the probability that they were born on different days? Because there is no * specified termination condition to terminate the recursion, a * StackOverflowError is to be expected. * * @return String variable. */ public String getStringVar() { // // http://canondrivebh.com/how-to/jbed-unfortunately-java-has-stopped.html How to deal with the StackOverflowError The simplest solution is to carefully inspect the stack trace and detect the repeating pattern of line numbers.

Not the answer you're looking for? Java.lang.stackoverflowerror Recursive Function The return address denotes the execution point from which, the program execution shall continue after the invoked method returns. A sample execution, using the -Xss1M flag that specifies the size of the thread stack to equal to 1MB, is shown below: Number: 1 Number: 2 Number: 3 ...

but not one that I understand. –Ziggy Oct 18 '08 at 9:04 3 Ha ha ha, so here it is: while (points < 100) {addMouseListeners(); moveball(); checkforcollision(); pause(speed);} Wow do

The arguments and the address of where the method was called is pushed on the stack (see http://en.wikipedia.org/wiki/Stack_(abstract_data_type)#Runtime_memory_management) so that the called method can access the arguments and so that when In Java, all direct allocations come from the heap, by using "new". –Chris Jester-Young Oct 18 '08 at 9:26 @ChrisJester-Young Isn't it true that if I have 100 local Try adding a -Xss argument (or increasing the value of one) to see if this goes away. Java.lang.stackoverflowerror Null The above example is somewhat simplified (although it happens to me more than I'd like to admit.) The same thing can happen in a more round about way making it a

So how can sort out this error ? –Ajay Sharma Jun 16 at 6:44 add a comment| up vote 16 down vote One of the (optional) arguments to the JVM is In this case, we can quickly realize that the intended behavior was to instead return this.stringVar;.Unintended Recursion with Cyclic RelationshipsThere are certain risks to having cyclic relationships between classes. The format of the -Xss argument is: -Xss[g|G|m|M|k|K] share|improve this answer edited Apr 24 '15 at 8:21 Mackan 4,04221029 answered Mar 26 '15 at 13:06 varun 1,28111337 add a comment| up Check This Out For more info see http://en.wikipedia.org/wiki/Recursion_(computer_science)) we are in a situation known as infinite recursion and we keep piling the arguments and return address on the call stack.

If the stack is full you can't push, if you do you'll get stack overflow error. A simple example is public static void main(String... return "StateName: " + this.name + NEW_LINE + "StateAbbreviation: " + this.abbreviation + NEW_LINE + "CapitalCity: " + this.capitalCity; } } City.java 1 2 Page 1 Next View Comments Recommended Eclipse, For me it's often when I intended to call a super method for the overidden method.

current community chat Stack Overflow Meta Stack Overflow your communities Sign up or log in to customize your list. Once you detect these lines, you must carefully inspect your code and understand why the recursion never terminates. args) { Main main = new Main(); main.testMethod(1); } public void testMethod(int i) { testMethod(i); System.out.println(i); } Here the System.out.println(i); will be repeatedly pushed to stack when the testMethod is called. In theory, it is also possible to have a stack overflow without recursion, but in practice, it would appear to be a fairly rare event.

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