Java : recursive constructor call and stackoverflow error

Question

Please help to understand why does the following code

public class HeapQn1 {

    /**
     * @param args
     */
    public HeapQn1() {
        new HeapQn1();
    }

    static HeapQn1 n = new HeapQn1();

    public static void main(String[] args) {

    }

}

results in

java.lang.StackOverflowError
    at com.rg.test.interview.HeapQn1.<init>(HeapQn1.java:8)
    at com.rg.test.interview.HeapQn1.<init>(HeapQn1.java:9)
    ...

As per my understanding the memory allocation for an object happens in the heap memory and I was expecting an OutOfMemoryError as at some point the heap memory will be full because of repetitive object creation.

On research , I came across that a java constructor is considered a method and that explained the StackOverflowError , until I read the following thread.

When does the Constructor gets called in java?

which says

3. The object is fully constructed/created when the constructor returns.

From what I could gather , the constructor is a method and since the heap memory is much larger than stack memory , the recursive constructor call resulted in StackOverflowError . Is this correct ?

Since no object in the give code will get completely created , will stack frame allocation for constructor actually happen ?

--edit-- For the duplicates pointed out , I do understand what StackoverflowError is . I have mentioned in the question "On research , I came across that a java constructor is considered a method and that explained the StackOverflowError". My question is to understand if a constructor gets a stack frame allocated just like other methods as the object creation is not complete until the constructor returns. Hope this clarifies.

Possible duplicate of stackoverflow.com/questions/214741/… — Raedwald
– Raedwald, Commented Jun 18, 2015 at 10:04
@Raedwald - I understand what a StackOverflowError is , my question is 1.how does why the code did not end up in OutOfMemoryError 2. is constructor a method . — R.G
– R.G, Commented Jun 18, 2015 at 10:09
I've made some empirical experiments and I've summarized them in my answer. Check it out :) — riccardo.cardin
– riccardo.cardin, Commented Jun 18, 2015 at 13:57

BaCaRoZzo · Accepted Answer · 2015-06-18 10:53:15Z

2

Whenever the constructor is called, its return address is pushed onto the stack. As the stack is finite and smaller than the heap memory, you are getting error like StackOverflowError rather than OutOfMemoryError.

The constructor is a method and since the heap memory is much larger than the stack memory, the recursive constructor call resulted in StackOverflowError. Is this correct ?

Yeah, your wild guess is completely correct. Cheers!

edited Jun 18, 2015 at 10:53

BaCaRoZzo

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answered Jun 18, 2015 at 9:48

mazhar islam

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Comments

riccardo.cardin · Accepted Answer · 2015-06-18 13:55:59Z

1

The constructor is a method, a.k.a. a function. Every time you call it a chunk of memory is allocated to the stack, to store the variables of the function.

Your code creates calls to the constructor function indefenitely, allocating memory to the stack until the memory finishes.

You are obtaining a StackOverflowError and not an OutOfMemoryError, because the quantity of memory dedicated to the stack is smaller than the quantity of memory dedicated to the heap.

EDIT: I have done some test using your code. I've specified a heap memory space of 8M (-Xms8M -Xmx8M) and a stack memory space of 100M (-Xss100M). The computation result is always the error StackOverflowError.

Then, this could mean that no memory is allocated to the heap in this case. As you stated in your question:

The object is fully constructed/created when the constructor returns.

edited Jun 18, 2015 at 13:55

answered Jun 18, 2015 at 9:44

riccardo.cardin

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7 Comments

Jagger Over a year ago

Isn't it the other way round?

riccardo.cardin Over a year ago

@Jagger Why are you saying that?

Jagger Over a year ago

You wrote that the memory consumed by the stack is greater than the one consumed by the heap. I think it should be the other way round.

riccardo.cardin Over a year ago

My english is not perfect, so it could be that I've wrote something wrong. The exception he obtained was a StackOverflowError, then the memory of the stack was consumed quicklier than the memory of the heap, wasn't it?

Jagger Over a year ago

Well, technically no, because if only the stack space runs out then you cannot even say of heap allocation in this case, can you?

|

Karrde · Accepted Answer · 2015-06-18 09:46:43Z

0

Basically what you said is correct, the stack space runs out before the heap space does.

answered Jun 18, 2015 at 9:46

Karrde

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Comments

khelwood · Accepted Answer · 2015-06-18 09:50:30Z

0

You are right: the stack is much smaller than the heap, and no object will be fully created.

edited Jun 18, 2015 at 9:50

khelwood

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answered Jun 18, 2015 at 9:42

Crazyjavahacking

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Comments

Lokesh Nirvan · Accepted Answer · 2018-12-22 16:38:15Z

0

As per the java all the reference variables are stored in the stack memory space and stack memory space is smaller than heap space that's what we get stackOverFlowException.

answered Dec 22, 2018 at 16:38

Lokesh Nirvan

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