This in Java - Exception in thread "main" java.lang.StackOverflowError

Question

Why am I getting a stackoverflow error ?

My class -

public class Tester {

int id;

 Tester(int id){
  this.id = id;
 }

 public String toString(){

  String rep = "Hex: " + this + ", Id: " + this.id;
  return rep;
 }

}

The main method -

class Driver{

    public static void main(String [] args){

        Tester t = new Tester(123);
        System.out.println(t);

    }

}

Error -

Exception in thread "main" java.lang.StackOverflowError
    at java.lang.String.length(Unknown Source)
    at java.lang.AbstractStringBuilder.append(Unknown Source)
    at java.lang.StringBuilder.append(Unknown Source)
    at java.lang.StringBuilder.<init>(Unknown Source)
    at com.examscam.model.Tester.toString(Tester.java:13)
    at java.lang.String.valueOf(Unknown Source)
    at java.lang.StringBuilder.append(Unknown Source)

---------REPEAT !!!

@VineetSingla - to show the value of `this`. But obviously, I can't do that without causing a SO error. — Erran Morad, Apr 28 '14 at 07:03
Printing this will just print the classname@hashcode values, Tester class has only one attribute id, so just printing the id should work. — Vineet Singla, Apr 28 '14 at 07:06
possible duplicate of [What is a stack trace, and how can I use it to debug my application errors?](http://stackoverflow.com/questions/3988788/what-is-a-stack-trace-and-how-can-i-use-it-to-debug-my-application-errors) — Raedwald, Apr 28 '14 at 07:24
Possible duplicate of http://stackoverflow.com/questions/214741/what-is-a-stack-overflow-error — Raedwald, Apr 28 '14 at 07:24

JB Nizet · Answer 1 · 2014-04-28T07:22:24.283

16

Because

"Hex: " + this

is equivalent to

"Hex: " + this.toString()

and you're doing that from the toString(), so toString() calls itself, which calls itself, which calls itself...

edited Apr 28 '14 at 07:22

answered Apr 28 '14 at 06:52

JB Nizet

678,734
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score 6 · Accepted Answer · answered Apr 28 '14 at 07:56

6

You wrote:

String rep = "Hex: " + this + ", Id: " + this.id;

In java simply writing this means that you are indirectly invoking this.toString().

I believe you are trying to override the toString() method of Object and inside your version of toString() you want to print the id you have passed along with the hashcode of the object.

So to get the output replace

String rep = "Hex: " + this + ", Id: " + this.id;

with

String rep = "Hex: "+ this.getClass().getName().hashCode() +", Id: " + id;

and you will get the output as:

Hex: 1800024669, Id: 123

answered Apr 28 '14 at 07:56

Parul S

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1

Thank you Parul. Yes, the oracle docs toString() have a same definitions. Borat very much like your answer ! High five ! – Erran Morad Apr 28 '14 at 17:14
You can also just use `super.toString()` to call the super class method. – Giovanni Botta Apr 29 '14 at 15:34

score 4 · Answer 3 · answered Apr 28 '14 at 06:52

4

you are using this keyword.

String rep = "Hex: " + this + ", Id: " + this.id;

This represent the current object. Your current object is being called again and again recursivley so you are getting

java.lang.StackOverflowError

answered Apr 28 '14 at 06:52

Rahul

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score 4 · Answer 4 · answered Apr 28 '14 at 06:52

4

You are appending "this". This calls the toString() method, which again calls toString(), ... It's an infinite recursion loop, which does not have an end.

answered Apr 28 '14 at 06:52

MrP

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score 4 · Answer 5 · answered Apr 28 '14 at 06:52

4

Because you are referencing this in toString()

That means that this.toString() is being called, therefor infinite recursion is occurring

answered Apr 28 '14 at 06:52

geoand

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score 2 · Answer 6 · answered Apr 28 '14 at 06:53

2

Your toString method is the culprit,

String rep = "Hex: " + super.toString() /* Not this */
   + ", Id: " + this.id;

answered Apr 28 '14 at 06:53

Elliott Frisch

198,278
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score 2 · Answer 7 · answered Apr 28 '14 at 06:54

2

thi line

String rep = "Hex: " + this + ", Id: " + this.id;

would become

String rep = "Hex: " + this.toString() + ", Id: " + this.id;

at run-time and will again call your class's toString..wi again..

answered Apr 28 '14 at 06:54

niiraj874u

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score 1 · Answer 8 · answered Apr 28 '14 at 06:58

1

In the line String rep = "Hex: " + this + ", Id: " + this.id;

  this

is equivalent to

 this.toString()

and calling it from the toString(), will again call toString and again...

answered Apr 28 '14 at 06:58

Vineet Singla

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score 0 · Answer 9 · answered Apr 28 '14 at 06:59

0

I think you are overriding toString method and in your overrided method body your calling your method again! you are calling toString by writing this+"" in toString method

answered Apr 28 '14 at 06:59

Blue Ocean

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score 0 · Answer 10 · answered Apr 28 '14 at 07:31

0

String rep = "Hex: " + this + ", Id: " + this.id;

equal to String rep = "Hex: " + this.toString() + ", Id: " + this.id; internally

this will leads to recursive method call that will result in

 java.lang.StackOverflowError

answered Apr 28 '14 at 07:31

Naveen

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score 0 · Answer 11 · answered Mar 04 '15 at 08:59

0

Because using first this in your toString method goes on calling the toString method recursively resulting in a stackoverflow.

answered Mar 04 '15 at 08:59

Abdullah Khan

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This in Java - Exception in thread "main" java.lang.StackOverflowError

11 Answers11