What is the maximum depth of the java call stack?

Question

How deep do I need to go into the call stack before I get a StackOverflowError? Is the answer platform dependent?

Closely related: http://stackoverflow.com/questions/794227/how-to-know-about-outofmemory-or-stackoverflow-errors-ahead-of-time — finnw, Jan 19 '11 at 10:35
Since this is a good question, I've updated the title to something I feel is more clearly associated with the meaning. (Previously I thought you might be referring to the depth of a *particular* stack you'd captured at runtime, for example). Feel free to change it back if you disagree. — Andrzej Doyle, Jan 19 '11 at 11:28

finnw · Accepted Answer · 2011-01-19T10:30:52.860

66

It depends on the amount of virtual memory allocated to the stack.

http://www.odi.ch/weblog/posting.php?posting=411

You can tune this with the -Xss VM parameter or with the Thread(ThreadGroup, Runnable, String, long) constructor.

edited Jan 19 '11 at 10:30

answered Jan 19 '11 at 10:25

finnw

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And perhaps the size of the stack frames you're putting on it? – duffymo Jan 19 '11 at 10:26
1

The limit also depends on the stack usage in the function. – Jarek Przygódzki May 04 '13 at 14:50
2

if we don't mention the Xss, then? – a3.14_Infinity Jul 04 '16 at 07:52

score 32 · Answer 2 · edited Aug 10 '16 at 08:47

32

I tested on my system and didn't find any constant value, sometimes stack overflow occurs after 8900 calls, sometimes only after 7700, random numbers.

public class MainClass {

    private static long depth=0L;

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

    private static void deep(){
        System.err.println(++depth);
        deep();
    }

}

edited Aug 10 '16 at 08:47

soufrk

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answered Jul 06 '14 at 08:29

troy

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Isn't it the case that this is tail recursive and shouldn't ever overflow? Edit: Sorry. In Java it crashed at 8027; in Scala it got up to 8594755 before I got bored. – arya Feb 15 '15 at 02:47
11

@arya an important part of the JVM semantics is that tail recursion is not supported. This gives a lot of interesting problems for those who want to implement languages with tail recursion on the JVM. – Thorbjørn Ravn Andersen Feb 20 '15 at 23:57
2

`public foo() { try { foo(); } finally { foo(); } }` can run 'virtually' forever, in java only though. – Felype Mar 23 '15 at 16:45
for me, `StackOverflowError` occurs after after 8792 – ericdemo07 Dec 05 '16 at 07:11
3

@ThorbjørnRavnAndersen tail recursion **optimisation** is not supported. Rather obviously you can have tail recursion. It just doesn't optimise it to not grow the call stack. – slim Aug 16 '17 at 14:48
@slim Of course, or it wouldn't be interesting. Feel free to write a full answer explaining this in detail. – Thorbjørn Ravn Andersen Aug 16 '17 at 14:53

score 20 · Answer 3 · answered Jan 19 '11 at 10:46

20

The stack size can be set with the -Xss command line switch but as a rule of thumb, it is deep enough, hundreds if not thousands of calls deep. (The default is platform dependent, but at least 256k in most platforms.)

If you get a stack overflow, 99% of the time it's caused by an error in the code.

answered Jan 19 '11 at 10:46

biziclop

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+1 for second paragraph. One should always remember about that. – mcveat Jan 19 '11 at 11:53
6

Using eclipse, I only get 1024 recursive calls. – Norswap Mar 25 '13 at 10:57
3

@Norswap Are you determining that by the size of the stack trace? That appears to be limited to 1024 regardless of the actual size of the stack. – Brian McCutchon Aug 17 '17 at 04:59

score 3 · Answer 4 · edited Aug 10 '16 at 08:06

Compare these two calls:
(1) Static method:

public static void main(String[] args) {
    int i = 14400; 
    while(true){   
        int myResult = testRecursion(i);
        System.out.println(myResult);
        i++;
    }
}

public static int testRecursion(int number) {
    if (number == 1) {
        return 1;
    } else {
        int result = 1 + testRecursion(number - 1);
        return result;
    }    
}
 //Exception in thread "main" java.lang.StackOverflowError after 62844

(2) Non-static method using a different class:

public static void main(String[] args) {
    int i = 14400;
    while(true){       
        TestRecursion tr = new TestRecursion ();
        int myResult = tr.testRecursion(i);
        System.out.println(myResult);
        i++;
    }
} 
//Exception in thread "main" java.lang.StackOverflowError after 14002

Test recursion class has public int testRecursion(int number) { as the only method.

score 0 · Answer 5 · answered Oct 29 '22 at 02:02

While traversing a tree with huge depth, we could face this


/**
 * -Xss10M
 * max call stack count 92766
 * -Xss100M
 * max call stack count 2148286
 */
public class MaxCallStack {
    private Integer stackCount = 0;

    public static void main(String[] args) {
        MaxCallStack maxStack = new MaxCallStack();
        maxStack.evaluate();
    }

    private void evaluate() {
        try {
            Tree tree = new Tree();
            tree.left = tree;
            traverse(tree);
        } catch (StackOverflowError e) {
            System.out.println(String.format("max call stack count %s", stackCount));
        }
    }

    private void traverse(Tree tree) {
        stackCount++;
        if (tree.left != null) {
            traverse(tree.left);
        }
        if (tree.right != null) {
            traverse(tree.right);
        }
    }

    private class Tree {
        Tree left;
        Tree right;
    }
}

What is the maximum depth of the java call stack?

5 Answers5

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