What is a StackOverflowError?

Question

What is a StackOverflowError, what causes it, and how should I deal with them?

Answer 1

The stack has a space limit that depends on the operating system. The normal size is 8 MB (in Ubuntu (Linux), you can check that limit with $ ulimit -u and it can be checked in other OS similarly). Any program makes use of the stack at runtime, but to fully know when it is used you need to check the assembly language. In x86_64 for example, the stack is used to:

1. Save the return address when making a procedure call
2. Save local variables
3. Save special registers to restore them later
4. Pass arguments to a procedure call (when there are more than 6 integers parameters and/or more than 8 floating-point parameters)
5. Other: random unused stack base, canary values, padding, ... etc.

If you don't know x86_64 (normal case) you only need to know when the specific high-level programming language you are using compile to those actions. For example in C:

• (1) → a function call
• (2) → local variables in function calls (including main)
• (3) → local variables in function calls (not main)
• (4) → a function call
• (5) → normally a function call, it is generally irrelevant for a stack overflow.

So, in C, only local variables and function calls make use of the stack. The two (unique?) ways of making a stack overflow are:

• Declaring too large local variables in main or in any function that it's called in (int array[10000][10000];)
• A very deep or infinite recursion (too many function calls at the same time).

To avoid a StackOverflowError you can:

• check if local variables are too big (order of 1 MB) → use the heap (malloc/calloc calls) or global variables.

• check for infinite recursion → you know what to do... correct it!

• check for normal too deep recursion → the easiest approach is to just change the implementation to be iterative.

Notice also that global variables, include libraries, etc... don't make use of the stack.

Only if the above does not work, change the stack size to the maximum on the specific OS. With Ubuntu for example: ulimit -s 32768 (32 MB). (This has never been the solution for any of my stack overflow errors, but I also don't have much experience.)

I have omitted special and/or not standard cases in C (such as usage of alloc() and similar) because if you are using them you should already know exactly what you are doing.

Answer 2

Many answers to this question are good. However, I would like to take a slightly different approach and give some more insight into how memory works and also a (simplified) visualization to better understand StackOverflow errors. This understanding does not only apply to Java but all processes alike.

On modern systems all new processes get their own virtual address space (VAS). In essence VAS is an abstraction layer provided by the operating system on top of physical memory in order to ensure processes do not interfere with each others memory. It's the kernels job to then map the virtual addresses provided to to the actual physical addresses.

---

VAS can be divided into a couple of sections:

In order to let the CPU know what it is supposed to do machine instructions must be loaded into memory. This is usually referred to as the code or text segment and of static size.

On top of that one can find the data segment and heap. The data segment is of fixed size and contains global or static variables. As a program runs into special conditions it may need to additionally allocate data, which is where the heap comes into play and is therefore able to dynamically grow in size.

The stack is located on the other side of the virtual address space and (among other things) keeps track of all function calls using a LIFO data structure. Similar to the heap a program may need additional space during runtime to keep track of new function calls being invoked. Since the stack is located on the other side of the VAS it is growing into the opposite direction i.e. towards the heap.

TL;DR

This is where the StackOverflow error comes into play.

Since the stack grows down (towards the heap) it may so happen that at some point in time it cannot grow further as it would overlap with the heap address space. Once that happens the StackOverflow error occurs.

The most common reason as to why this happens is due to a bug in the program making recursive calls that do not terminate properly.

---

Note that on some systems VAS may behave slightly different an can be divided into even more segments, however, this general understanding applies to all UNIX systems.

Answer 3

A StackOverflowError is a runtime error in Java.

It is thrown when the amount of call stack memory allocated by the JVM is exceeded.

A common case of a StackOverflowError being thrown, is when the call stack exceeds due to excessive deep or infinite recursion.

Example:

public class Factorial {
    public static int factorial(int n){
        if(n == 1){
            return 1;
        }
        else{
            return n * factorial(n-1);
        }
    }

    public static void main(String[] args){
         System.out.println("Main method started");
        int result = Factorial.factorial(-1);
        System.out.println("Factorial ==>"+result);
        System.out.println("Main method ended");
    }
}

Stack trace:

Main method started
Exception in thread "main" java.lang.StackOverflowError
at com.program.stackoverflow.Factorial.factorial(Factorial.java:9)
at com.program.stackoverflow.Factorial.factorial(Factorial.java:9)
at com.program.stackoverflow.Factorial.factorial(Factorial.java:9)

In the above case, it can be avoided by doing programmatic changes. But if the program logic is correct and it still occurs then your stack size needs to be increased.

Answer 4

Here is an example of a recursive algorithm for reversing a singly linked list. On a laptop (with the specifications 4 GB memory, Intel Core i5 2.3 GHz CPU 64 bit and Windows 7), this function will run into StackOverflow error for a linked list of size close to 10,000.

My point is that we should use recursion judiciously, always taking into account of the scale of the system.

Often recursion can be converted to iterative program, which scales better. (One iterative version of the same algorithm is given at the bottom of the page. It reverses a singly linked list of size 1 million in 9 milliseconds.)

private static LinkedListNode doReverseRecursively(LinkedListNode x, LinkedListNode first){

    LinkedListNode second = first.next;

    first.next = x;

    if(second != null){
        return doReverseRecursively(first, second);
    }else{
        return first;
    }
}


public static LinkedListNode reverseRecursively(LinkedListNode head){
    return doReverseRecursively(null, head);
}

Iterative Version of the Same Algorithm:

public static LinkedListNode reverseIteratively(LinkedListNode head){
    return doReverseIteratively(null, head);
}


private static LinkedListNode doReverseIteratively(LinkedListNode x, LinkedListNode first) {

    while (first != null) {
        LinkedListNode second = first.next;
        first.next = x;
        x = first;

        if (second == null) {
            break;
        } else {
            first = second;
        }
    }
    return first;
}


public static LinkedListNode reverseIteratively(LinkedListNode head){
    return doReverseIteratively(null, head);
}

Answer 5

This is a typical case of java.lang.StackOverflowError... The method is recursively calling itself with no exit in doubleValue(), floatValue(), etc.

File Rational.java

public class Rational extends Number implements Comparable<Rational> {
    private int num;
    private int denom;

    public Rational(int num, int denom) {
        this.num = num;
        this.denom = denom;
    }

    public int compareTo(Rational r) {
        if ((num / denom) - (r.num / r.denom) > 0) {
            return +1;
        } else if ((num / denom) - (r.num / r.denom) < 0) {
            return -1;
        }
        return 0;
    }

    public Rational add(Rational r) {
        return new Rational(num + r.num, denom + r.denom);
    }

    public Rational sub(Rational r) {
        return new Rational(num - r.num, denom - r.denom);
    }

    public Rational mul(Rational r) {
        return new Rational(num * r.num, denom * r.denom);
    }

    public Rational div(Rational r) {
        return new Rational(num * r.denom, denom * r.num);
    }

    public int gcd(Rational r) {
        int i = 1;
        while (i != 0) {
            i = denom % r.denom;
            denom = r.denom;
            r.denom = i;
        }
        return denom;
    }

    public String toString() {
        String a = num + "/" + denom;
        return a;
    }

    public double doubleValue() {
        return (double) doubleValue();
    }

    public float floatValue() {
        return (float) floatValue();
    }

    public int intValue() {
        return (int) intValue();
    }

    public long longValue() {
        return (long) longValue();
    }
}

File Main.java

public class Main {

    public static void main(String[] args) {

        Rational a = new Rational(2, 4);
        Rational b = new Rational(2, 6);

        System.out.println(a + " + " + b + " = " + a.add(b));
        System.out.println(a + " - " + b + " = " + a.sub(b));
        System.out.println(a + " * " + b + " = " + a.mul(b));
        System.out.println(a + " / " + b + " = " + a.div(b));

        Rational[] arr = {new Rational(7, 1), new Rational(6, 1),
                new Rational(5, 1), new Rational(4, 1),
                new Rational(3, 1), new Rational(2, 1),
                new Rational(1, 1), new Rational(1, 2),
                new Rational(1, 3), new Rational(1, 4),
                new Rational(1, 5), new Rational(1, 6),
                new Rational(1, 7), new Rational(1, 8),
                new Rational(1, 9), new Rational(0, 1)};

        selectSort(arr);

        for (int i = 0; i < arr.length - 1; ++i) {
            if (arr[i].compareTo(arr[i + 1]) > 0) {
                System.exit(1);
            }
        }


        Number n = new Rational(3, 2);

        System.out.println(n.doubleValue());
        System.out.println(n.floatValue());
        System.out.println(n.intValue());
        System.out.println(n.longValue());
    }

    public static <T extends Comparable<? super T>> void selectSort(T[] array) {

        T temp;
        int mini;

        for (int i = 0; i < array.length - 1; ++i) {

            mini = i;

            for (int j = i + 1; j < array.length; ++j) {
                if (array[j].compareTo(array[mini]) < 0) {
                    mini = j;
                }
            }

            if (i != mini) {
                temp = array[i];
                array[i] = array[mini];
                array[mini] = temp;
            }
        }
    }
}

Result

2/4 + 2/6 = 4/10
Exception in thread "main" java.lang.StackOverflowError
2/4 - 2/6 = 0/-2
    at com.xetrasu.Rational.doubleValue(Rational.java:64)
2/4 * 2/6 = 4/24
    at com.xetrasu.Rational.doubleValue(Rational.java:64)
2/4 / 2/6 = 12/8
    at com.xetrasu.Rational.doubleValue(Rational.java:64)
    at com.xetrasu.Rational.doubleValue(Rational.java:64)
    at com.xetrasu.Rational.doubleValue(Rational.java:64)
    at com.xetrasu.Rational.doubleValue(Rational.java:64)
    at com.xetrasu.Rational.doubleValue(Rational.java:64)

Here is the source code of StackOverflowError in OpenJDK 7.

Answer 6

In a crunch, the below situation will bring a stack overflow error.

public class Example3 {

    public static void main(String[] args) {

        main(new String[1]);
    }

}

Answer 7

A simple Java example that causes java.lang.StackOverflowError because of a bad recursive call:

class Human {
    Human(){
        new Animal();
    }
}

class Animal extends Human {
    Animal(){
        super();
    }
}

public class Test01 {
    public static void main(String[] args) {
        new Animal();
    }
}

Answer 8

Parameters and local variables are allocated on the stack (with reference types, the object lives on the heap and a variable in the stack references that object on the heap). The stack typically lives at the upper end of your address space and as it is used up it heads towards the bottom of the address space (i.e. towards zero).

Your process also has a heap, which lives at the bottom end of your process. As you allocate memory, this heap can grow towards the upper end of your address space. As you can see, there is a potential for the heap to "collide" with the stack (a bit like tectonic plates!!!).

The common cause for a stack overflow is a bad recursive call. Typically, this is caused when your recursive functions doesn't have the correct termination condition, so it ends up calling itself forever. Or when the termination condition is fine, it can be caused by requiring too many recursive calls before fulfilling it.

However, with GUI programming, it's possible to generate indirect recursion. 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. Here you've not explicitly called yourself, but the OS/VM has done it for you.

To deal with them, you'll need to examine your code. If you've got functions that call themselves then check that you've got a terminating condition. If you have, then check that when calling the function you have at least modified one of the arguments, otherwise there'll be no visible change for the recursively called function and the terminating condition is useless. Also mind that your stack space can run out of memory before reaching a valid terminating condition, thus make sure your method can handle input values requiring more recursive calls.

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).

Answer 9

StackOverflowError is to the stack as OutOfMemoryError is to the heap.

Unbounded recursive calls result in stack space being used up.

The following example produces StackOverflowError:

class  StackOverflowDemo
{
    public static void unboundedRecursiveCall() {
     unboundedRecursiveCall();
    }

    public static void main(String[] args) 
    {
        unboundedRecursiveCall();
    }
}

StackOverflowError is avoidable if recursive calls are bounded to prevent the aggregate total of incomplete in-memory calls (in bytes) from exceeding the stack size (in bytes).

Answer 10

Here's an example

public static void main(String[] args) {
    System.out.println(add5(1));
}

public static int add5(int a) {
    return add5(a) + 5;
}

A StackOverflowError basically is when you try to do something, that most likely calls itself, and goes on for infinity (or until it gives a StackOverflowError).

add5(a) will call itself, and then call itself again, and so on.

Answer 11

Stack overflow means exactly that: a stack overflows. Usually there's a one stack in the program that contains local-scope variables and addresses where to return when execution of a routine ends. That stack tends to be a fixed memory range somewhere in the memory, therefore it's limited how much it can contain values.

If the stack is empty you can't pop, if you do you'll get stack underflow error.

If the stack is full you can't push, if you do you'll get stack overflow error.

So stack overflow appears where you allocate too much into the stack. For instance, in the mentioned recursion.

Some implementations optimize out some forms of recursions. Tail recursion in particular. Tail recursive routines are form of routines where the recursive call appears as a final thing what the routine does. Such routine call gets simply reduced into a jump.

Some implementations go so far as implement their own stacks for recursion, therefore they allow the recursion to continue until the system runs out of memory.

Easiest thing you could try would be to increase your stack size if you can. 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. Try it by printing something before and after the call into routine. This helps you to find out the failing routine.

Answer 12

The most common cause of stack overflows is excessively deep or infinite recursion. If this is your problem, this tutorial about Java Recursion could help understand the problem.

Answer 13

If you have a function like:

int foo()
{
    // more stuff
    foo();
}

Then foo() will keep calling itself, getting deeper and deeper, and when the space used to keep track of what functions you're in is filled up, you get the stack overflow error.

Answer 14

A stack overflow is usually called by nesting function calls too deeply (especially easy when using recursion, i.e. a function that calls itself) or allocating a large amount of memory on the stack where using the heap would be more appropriate.

Answer 15

Like you say, you need to show some code. :-)

A stack overflow error usually happens when your function calls nest too deeply. See the Stack Overflow Code Golf thread for some examples of how this happens (though in the case of that question, the answers intentionally cause stack overflow).