Ошибка переполнения стека java

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.

Aren’t there other ways for a stack overflow to occur, not only
through recursion?

Challenge accepted :) StackOverflowError without recursion (challenge failed, see comments):

public class Test
{
    final static int CALLS = 710;

    public static void main(String[] args)
    {
        final Functor[] functors = new Functor[CALLS];
        for (int i = 0; i < CALLS; i++)
        {
            final int finalInt = i;
            functors[i] = new Functor()
            {
                @Override
                public void fun()
                {
                    System.out.print(finalInt + " ");
                    if (finalInt != CALLS - 1)
                    {
                        functors[finalInt + 1].fun();
                    }
                }
            };
        }
        // Let's get ready to ruuuuuuumble!
        functors[0].fun(); // Sorry, couldn't resist to not comment in such moment. 
    }

    interface Functor
    {
        void fun();
    }
}

Compile with standard javac Test.java and run with java -Xss104k Test 2> out. After that, more out will tell you:

Exception in thread "main" java.lang.StackOverflowError

Second try.

Now the idea is even simpler. Primitives in Java can be stored on the stack. So, let’s declare a lot of doubles, like double a1,a2,a3.... This script can write, compile and run the code for us:

#!/bin/sh

VARIABLES=4000
NAME=Test
FILE=$NAME.java
SOURCE="public class $NAME{public static void main(String[] args){double "
for i in $(seq 1 $VARIABLES);
do
    SOURCE=$SOURCE"a$i,"
done
SOURCE=$SOURCE"b=0;System.out.println(b);}}"
echo $SOURCE > $FILE
javac $FILE
java -Xss104k $NAME

And… I got something unexpected:

#
# A fatal error has been detected by the Java Runtime Environment:
#
#  SIGSEGV (0xb) at pc=0x00007f4822f9d501, pid=4988, tid=139947823249152
#
# JRE version: 6.0_27-b27
# Java VM: OpenJDK 64-Bit Server VM (20.0-b12 mixed mode linux-amd64 compressed oops)
# Derivative: IcedTea6 1.12.6
# Distribution: Ubuntu 10.04.1 LTS, package 6b27-1.12.6-1ubuntu0.10.04.2
# Problematic frame:
# V  [libjvm.so+0x4ce501]  JavaThread::last_frame()+0xa1
#
# An error report file with more information is saved as:
# /home/adam/Desktop/test/hs_err_pid4988.log
#
# If you would like to submit a bug report, please include
# instructions how to reproduce the bug and visit:
#   https://bugs.launchpad.net/ubuntu/+source/openjdk-6/
#
Aborted

It’s 100% repetitive. This is related to your second question:

Does the StackOverflowError happen before the JVM actually overflows
the stack or after?

So, in case of OpenJDK 20.0-b12 we can see that JVM firstly exploded. But it seems like a bug, maybe someone can confirm that in comments please, because I’m not sure. Should I report this? Maybe it’s already fixed in some newer version… According to JVM specification link (given by JB Nizet in a comment) JVM should throw a StackOverflowError, not die:

If the computation in a thread requires a larger Java Virtual Machine
stack than is permitted, the Java Virtual Machine throws a
StackOverflowError.


Third try.

public class Test {
    Test test = new Test();

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

We want to create new Test object. So, its (implicit) constructor will be called. But, just before that, all the members of Test are initialized. So, Test test = new Test() is executed first…

We want to create new Test object…

Update: Bad luck, this is recursion, I asked question about that here.

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    StackOverflowError is an error which Java doesn’t allow to catch, for instance, stack running out of space, as it’s one of the most common runtime errors one can encounter.

    The main cause of the StackOverflowError is that we haven’t provided the proper terminating condition to our recursive function or template, which means it will turn into an infinite loop.

    When does StackOverflowError encountered?

    When we invoke a method, a new stack frame is created on the call stack or on the thread stack size. This stack frame holds parameters of the invoked method, mostly the local variables and the return address of the method. The creation of these stack frames will be iterative and will be stopped only when the end of the method invokes is found in the nested methods. In amidst of this process, if JVM runs out of space for the new stack frames which are required to be created, it will throw a StackOverflowError.

    For example: Lack of proper or no termination condition. This is mostly the cause of this situation termed as unterminated or infinite recursion.

    Given below is the implementation of infinite recursion:

    public class StackOverflowErrorClass {

        static int i = 0;

        public static int printNumber(int x)

        {

            i = i + 2;

            System.out.println(i);

            return i + printNumber(i + 2);

        }

        public static void main(String[] args)

        {

            StackOverflowErrorClass.printNumber(i);

        }

    }

    Runtime Error:

    RunTime Error in java code :- Exception in thread “main” java.lang.StackOverflowError
    at java.io.PrintStream.write(PrintStream.java:526)
    at java.io.PrintStream.print(PrintStream.java:597)
    at java.io.PrintStream.println(PrintStream.java:736)
    at StackOverflowErrorClass.printNumber(StackOverflowErrorClass.java:13)
    at StackOverflowErrorClass.printNumber(StackOverflowErrorClass.java:14)
    at StackOverflowErrorClass.printNumber(StackOverflowErrorClass.java:14)
    .
    .
    .

    Note: Please run this on your system to see an error thrown, due to the stack size, this may not show error on online IDE.

    How to fix StackOverflowError?

    1. Avoiding repetitive calls: Try to introduce a proper terminating condition or some condition for the recursive calls to ensure that it terminates.

      Given below is the implementation with proper terminating condition:

      public class stackOverflow {

          static int i = 0;

          public static int printNumber(int x)

          {

              i = i + 2;

              System.out.println(i);

              if (i == 10)

                  return i;

              return i + printNumber(i + 2);

          }

          public static void main(String[] args)

          {

              stackOverflow.printNumber(i);

          }

      }

    2. Increasing the Stack Size: The second method could be, if you notice that it’s implemented correctly still we see an error, then we can avoid that only by increasing the Stack Size in order to store the required number of recursive calls. This is achieved by changing the settings of the compiler.

      Cyclic Relationships between classes is the relationship caused when two different classes instantiate each other inside their constructors.

      StackOverflowError is encountered because the constructor of Class A1 is instantiating Class A2, and the constructor of Class A2 is again instantiating Class A1, and it occurs repeatedly until we see StackOverflow. This error is mainly due to the bad calling of constructors, that is, calling each other, which is not even required, and also it doesn’t hold any significance, so we can just avoid introducing them in the codes.

      Given below is the implementation of Cyclic Relationships Between Classes:

      public class A1 {

          public A2 type2;

          public A1()

          {

              type2 = new A2();

          }

          public static void main(String[] args)

          {

              A1 type1 = new A1();

          }

      }

      class A2 {

          public A1 type1;

          public A2()

          {

              type1 = new A1();

          }

      }

      Runtime Error:

      RunTime Error in java code :- Exception in thread “main” java.lang.StackOverflowError
      at A2.(A1.java:32)
      at A1.(A1.java:13)
      at A2.(A1.java:32)
      .
      .
      .

      Note: This will keep repeating, Infinite Recursion is seen by these infinite cyclic calls.

    Last Updated :
    07 Apr, 2020

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    The java.lang.StackOverflowError is a runtime error which points to serious problems that cannot be caught by an application. The java.lang.StackOverflowError indicates that the application stack is exhausted and is usually caused by deep or infinite recursion.

    What Causes java.lang.StackOverflowError in Java

    The java.lang.StackOverflowError occurs when the application stack continues to grow until it reaches the maximum limit. Some of the most common causes for a java.lang.StackOverflowError are:

    1. Deep or infinite recursion — If a method calls itself recursively without a terminating condition.
    2. Cyclic relationships between classes — If a class A instantiates an object of class B, which in turn instantiates an object of class A. This can be considered as a form of recursion.
    3. Memory intensive applications — Applications that rely on resource heavy objects such as XML documents, GUI or java2D classes.

    java.lang.StackOverflowError Example in Java

    Here is an example of java.lang.StackOverflowError thrown due to unintended recursion:

    public class StackOverflowErrorExample {
        public void print(int myInt) {
            System.out.println(myInt);
            print(myInt);
        }
    
        public static void main(String[] args) {
            StackOverflowErrorExample soee = new StackOverflowErrorExample();
            soee.print(0);
        }
    }
    

    In this example, the recursive method print() calls itself over and over again until it reaches the maximum size of the Java thread stack since a terminating condition is not provided for the recursive calls. When the maximum size of the stack is reached, the program exits with a java.lang.StackOverflowError:

    Exception in thread "main" java.lang.StackOverflowError
        at java.base/sun.nio.cs.UTF_8$Encoder.encodeLoop(UTF_8.java:564)
        at java.base/java.nio.charset.CharsetEncoder.encode(CharsetEncoder.java:585)
        at java.base/sun.nio.cs.StreamEncoder.implWrite(StreamEncoder.java:301)
        at java.base/sun.nio.cs.StreamEncoder.implWrite(StreamEncoder.java:290)
        at java.base/sun.nio.cs.StreamEncoder.write(StreamEncoder.java:131)
        at java.base/java.io.OutputStreamWriter.write(OutputStreamWriter.java:208)
        at java.base/java.io.BufferedWriter.flushBuffer(BufferedWriter.java:120)
        at java.base/java.io.PrintStream.writeln(PrintStream.java:722)
        at java.base/java.io.PrintStream.println(PrintStream.java:938)
        at StackOverflowErrorExample.print(StackOverflowErrorExample.java:3)
        at StackOverflowErrorExample.print(StackOverflowErrorExample.java:4)
        at StackOverflowErrorExample.print(StackOverflowErrorExample.java:4)
        at StackOverflowErrorExample.print(StackOverflowErrorExample.java:4)
        at StackOverflowErrorExample.print(StackOverflowErrorExample.java:4)
    

    How to fix java.lang.StackOverflowError in Java

    Inspect the stack trace

    Carefully inspecting the error stack trace and looking for the repeating pattern of line numbers enables locating the line of code with the recursive calls. When the line is identified, the code should be examined and fixed by specifying a proper terminating condition. As an example, the error stack trace seen earlier can be inspected:

    at java.base/java.io.PrintStream.writeln(PrintStream.java:722)
    at java.base/java.io.PrintStream.println(PrintStream.java:938)
    at StackOverflowErrorExample.print(StackOverflowErrorExample.java:3)
    at StackOverflowErrorExample.print(StackOverflowErrorExample.java:4)
    at StackOverflowErrorExample.print(StackOverflowErrorExample.java:4)
    at StackOverflowErrorExample.print(StackOverflowErrorExample.java:4)
    at StackOverflowErrorExample.print(StackOverflowErrorExample.java:4)

    In the above trace, line number 4 can be seen repeating, which is where the recursive calls are made and causing java.lang.StackOverflowError.

    Increase Thread Stack Size (-Xss)

    If the code has been updated to implement correct recursion and the program still throws a java.lang.StackOverflowError, the thread stack size can be increased to allow a larger number of invocations. Increasing the stack size can be useful, for example, when the program involves calling a large number of methods or using lots of local variables.

    The stack size can be increased by changing the -Xss argument on the JVM, which can be set when starting the application. Here is an example:

    -Xss4m

    This will set the thread’s stack size to 4 mb which should prevent the JVM from throwing a java.lang.StackOverflowError.

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    Introduction

    StackOverFlowError is one of the common JVM error. In this blog post, let’s learn inner mechanics of thread stacks, reasons that can trigger StackOverFlowError and potential solutions to address this error.

    To gain a deeper understanding of this exception, let’s review this simple program:

    public class SimpleExample {
    
     public static void main(String args[]) {
    
      a();
     }
    
     public static void a() {
    
      int x = 0;
      b();
     }
    
     public static void b() {
    
      Car y = new Car();
      c();
     }
    
     public static void c() {
    
      float z = 0 f;
      System.out.println("Hello");
     }
    }

    This program is very simple with the following execution code:

    • main() method invoked first
    • main() method invokes a() method. Inside a() method integer variable ‘x’ is initialized to value 0.
    • a() method in turn invokes b() method. Inside b() method Car object constructed and assigned to variable ‘y’.
    • b() method in turn invokes c() method. Inside c() method float variable ‘z’ is initialized to value 0.

    Now let’s review what happens behind the scenes when the above simple program executed. Each thread in the application has its own stack. Each stack has multiple stack frames. Thread adds the methods it’s executing, primitive data types, object pointers, return values to its stack frame in the sequence order in which they got executed.

    thread-stack-frame

    In step #1: main() method pushed into the application thread’s stack. Step #2: a() method pushed into application thread’s stack. In a() method, primitive data type ‘int’ is defined with value 0 and assigned to variable x. This information also pushed into the same stack frame. Note both data i.e. ‘0’ and variable ‘x’ pushed into thread’s stack frame.

    In step #3: b() method pushed into thread’s stack. In b() method, ‘Car’ object created and assigned to variable ‘y’. A crucial point to note here is ‘Car’ object created in the heap and not in the thread’s stack. Only Car object’s reference i.e. y stored in the thread’s stack frame.

    In step #4: c() method pushed into thread’s stack. In c() method, primitive data type ‘float’ is defined with value 0f and assigned to variable z. This information is also pushed into the same stack frame. Note both data i.e. ‘0f’ and variable ‘z’ is pushed into thread’s stack frame.

    thread-stack-frame

    2. StackOverFlowError Cause

    As you can see thread’s stack is storing methods it’s executing, primitive data types, variables, object pointers and return values. All of these consume memory. If thread’s stack sizes grow beyond the allocated memory limit then StackOverflowError is thrown. Let’s look at the below buggy program, which will result in StackOverflowError:

    public class SOFDemo {
    
     public static void a() {
    
      // Buggy line. It will cause method a() to be called infinite number of times.
      a();
     }
    
     public static void main(String args[]) {
    
      a();
     }
    }

    In this program main() method invokes a() method. a() method recursively calls itself. This implementation will cause a() method to be invoked an infinite number of times. In this circumstance, a() method got added to thread’s stack frame an infinite number of times. Thus, after a few thousand iterations thread’s stack size limit might exceeded. Once stack size limit exceeds, it will result in ‘StackOverflowError’:

    Exception in thread "main" java.lang.StackOverflowError
    	at com.buggyapp.stackoverflow.SOFDemo.a(SOFDemo.java:7)
    	at com.buggyapp.stackoverflow.SOFDemo.a(SOFDemo.java:7)
    	at com.buggyapp.stackoverflow.SOFDemo.a(SOFDemo.java:7)
    	at com.buggyapp.stackoverflow.SOFDemo.a(SOFDemo.java:7)
    	at com.buggyapp.stackoverflow.SOFDemo.a(SOFDemo.java:7)
    	at com.buggyapp.stackoverflow.SOFDemo.a(SOFDemo.java:7)
    	at com.buggyapp.stackoverflow.SOFDemo.a(SOFDemo.java:7)
    	at com.buggyapp.stackoverflow.SOFDemo.a(SOFDemo.java:7)
    

    StackOverflowError

    3. Solutions to StackOverflowError

    There are a couple of strategies to address StackOverflowError.

    3.1 Fix the Code

    Because of a non-terminating recursive call (as shown in the above example), threads stack size can grow to a large size. In that circumstance, you must fix the source code which is causing recursive looping. When ‘StackOverflowError’ is thrown, it will print the stacktrace of the code that it was recursively executing. This code is a good pointer to start debugging and fixing the issue. In the above example it’s ‘a()’  method.

    3.2 Increase Thread Stack Size (-Xss)

    There might be a legitimate reason where a threads stack size needs an increment. Maybe thread has to execute a large number of methods or lot of local variables/created in the methods thread has been executing. In such circumstance, you can increase the thread’s stack size using the JVM argument: ‘-Xss’. Pass this argument when you start the application. Example:

    -Xss2m

    This will set the thread’s stack size to 2 Mb. It might bring a question, what is the default thread’s stack size? Default thread stack size varies based on your operating system, java version & vendor.

    JVM version

    Thread stack size

      Sparc 32-bit JVM   

    512k

      Sparc 64-bit JVM  

    1024k

      x86 Solaris/Linux 32-bit JVM

    320K

      x86 Solaris/Linux 64-bit JVM

    1024K

      Windows 32-bit JVM

    320K

      Windows 64-bit JVM

    1024K

    Summary

    In this article, we took a closer look at the StackOverflowError.We got an understanding of the cause of this error along with the options to fix this issue.

    Автор оригинала: baeldung.

    1. Обзор

    StackOverflowError может раздражать разработчиков Java, так как это одна из самых распространенных ошибок во время выполнения, с которой мы можем столкнуться.

    В этой статье мы рассмотрим, как может возникнуть эта ошибка, рассмотрев различные примеры кода, а также то, как мы можем с ней справиться.

    Давайте начнем с основ. При вызове метода в стеке вызовов создается новый кадр стека. Этот кадр стека содержит параметры вызываемого метода, его локальные переменные и адрес возврата метода, т. е. точку, из которой выполнение метода должно продолжаться после возврата вызванного метода.

    Создание фреймов стека будет продолжаться до тех пор, пока не будет достигнут конец вызовов методов, найденных внутри вложенных методов.

    Во время этого процесса, если JVM столкнется с ситуацией, когда нет места для создания нового кадра стека, он выдаст StackOverflowError .

    Наиболее распространенной причиной, по которой JVM сталкивается с этой ситуацией, является unterminated/бесконечная рекурсия – в описании Javadoc для StackOverflowError упоминается, что ошибка возникает в результате слишком глубокой рекурсии в конкретном фрагменте кода.

    Однако рекурсия-не единственная причина этой ошибки. Это также может произойти в ситуации, когда приложение продолжает вызывать методы из методов до тех пор, пока стек не будет исчерпан . Это редкий случай, поскольку ни один разработчик не будет намеренно следовать плохим методам кодирования. Другой редкой причиной является наличие огромного количества локальных переменных внутри метода .

    Ошибка StackOverflowError также может быть вызвана, когда приложение предназначено для циклических отношений c между классами . В этой ситуации конструкторы друг друга вызываются повторно, что приводит к возникновению этой ошибки. Это также можно рассматривать как форму рекурсии.

    Другой интересный сценарий, который вызывает эту ошибку, заключается в том, что экземпляр класса создается в том же классе, что и переменная экземпляра этого класса . Это приведет к тому, что конструктор одного и того же класса будет вызываться снова и снова (рекурсивно), что в конечном итоге приведет к ошибке StackOverflowError.

    В следующем разделе мы рассмотрим некоторые примеры кода, демонстрирующие эти сценарии.

    3. Ошибка StackOverflowError в действии

    В примере, показанном ниже, StackOverflowError будет вызван из-за непреднамеренной рекурсии, когда разработчик забыл указать условие завершения для рекурсивного поведения:

    public class UnintendedInfiniteRecursion {
        public int calculateFactorial(int number) {
            return number * calculateFactorial(number - 1);
        }
    }

    Здесь ошибка возникает во всех случаях для любого значения, переданного в метод:

    public class UnintendedInfiniteRecursionManualTest {
        @Test(expected = StackOverflowError.class)
        public void givenPositiveIntNoOne_whenCalFact_thenThrowsException() {
            int numToCalcFactorial= 1;
            UnintendedInfiniteRecursion uir 
              = new UnintendedInfiniteRecursion();
            
            uir.calculateFactorial(numToCalcFactorial);
        }
        
        @Test(expected = StackOverflowError.class)
        public void givenPositiveIntGtOne_whenCalcFact_thenThrowsException() {
            int numToCalcFactorial= 2;
            UnintendedInfiniteRecursion uir 
              = new UnintendedInfiniteRecursion();
            
            uir.calculateFactorial(numToCalcFactorial);
        }
        
        @Test(expected = StackOverflowError.class)
        public void givenNegativeInt_whenCalcFact_thenThrowsException() {
            int numToCalcFactorial= -1;
            UnintendedInfiniteRecursion uir 
              = new UnintendedInfiniteRecursion();
            
            uir.calculateFactorial(numToCalcFactorial);
        }
    }

    Однако в следующем примере указано условие завершения, но оно никогда не выполняется, если значение -1 передается в метод calculateFactorial () , который вызывает нескончаемую/бесконечную рекурсию:

    public class InfiniteRecursionWithTerminationCondition {
        public int calculateFactorial(int number) {
           return number == 1 ? 1 : number * calculateFactorial(number - 1);
        }
    }

    Этот набор тестов демонстрирует этот сценарий:

    public class InfiniteRecursionWithTerminationConditionManualTest {
        @Test
        public void givenPositiveIntNoOne_whenCalcFact_thenCorrectlyCalc() {
            int numToCalcFactorial = 1;
            InfiniteRecursionWithTerminationCondition irtc 
              = new InfiniteRecursionWithTerminationCondition();
    
            assertEquals(1, irtc.calculateFactorial(numToCalcFactorial));
        }
    
        @Test
        public void givenPositiveIntGtOne_whenCalcFact_thenCorrectlyCalc() {
            int numToCalcFactorial = 5;
            InfiniteRecursionWithTerminationCondition irtc 
              = new InfiniteRecursionWithTerminationCondition();
    
            assertEquals(120, irtc.calculateFactorial(numToCalcFactorial));
        }
    
        @Test(expected = StackOverflowError.class)
        public void givenNegativeInt_whenCalcFact_thenThrowsException() {
            int numToCalcFactorial = -1;
            InfiniteRecursionWithTerminationCondition irtc 
              = new InfiniteRecursionWithTerminationCondition();
    
            irtc.calculateFactorial(numToCalcFactorial);
        }
    }

    В данном конкретном случае ошибки можно было бы полностью избежать, если бы условие завершения было просто сформулировано как:

    public class RecursionWithCorrectTerminationCondition {
        public int calculateFactorial(int number) {
            return number <= 1 ? 1 : number * calculateFactorial(number - 1);
        }
    }

    Вот тест, который показывает этот сценарий на практике:

    public class RecursionWithCorrectTerminationConditionManualTest {
        @Test
        public void givenNegativeInt_whenCalcFact_thenCorrectlyCalc() {
            int numToCalcFactorial = -1;
            RecursionWithCorrectTerminationCondition rctc 
              = new RecursionWithCorrectTerminationCondition();
    
            assertEquals(1, rctc.calculateFactorial(numToCalcFactorial));
        }
    }

    Теперь давайте рассмотрим сценарий, в котором StackOverflowError возникает в результате циклических отношений между классами. Давайте рассмотрим ClassOne и classstwo , которые создают экземпляры друг друга внутри своих конструкторов, вызывая циклическую связь:

    public class ClassOne {
        private int oneValue;
        private ClassTwo clsTwoInstance = null;
        
        public ClassOne() {
            oneValue = 0;
            clsTwoInstance = new ClassTwo();
        }
        
        public ClassOne(int oneValue, ClassTwo clsTwoInstance) {
            this.oneValue = oneValue;
            this.clsTwoInstance = clsTwoInstance;
        }
    }
    public class ClassTwo {
        private int twoValue;
        private ClassOne clsOneInstance = null;
        
        public ClassTwo() {
            twoValue = 10;
            clsOneInstance = new ClassOne();
        }
        
        public ClassTwo(int twoValue, ClassOne clsOneInstance) {
            this.twoValue = twoValue;
            this.clsOneInstance = clsOneInstance;
        }
    }

    Теперь предположим, что мы попытаемся создать экземпляр Class One , как показано в этом тесте:

    public class CyclicDependancyManualTest {
        @Test(expected = StackOverflowError.class)
        public void whenInstanciatingClassOne_thenThrowsException() {
            ClassOne obj = new ClassOne();
        }
    }

    Это заканчивается StackOverflowError , так как конструктор Class One создает экземпляр Class Two, и конструктор classwo снова создает экземпляр ClassOne. И это повторяется до тех пор, пока он не переполнит стек.

    Далее мы рассмотрим, что происходит, когда экземпляр класса создается в том же классе, что и переменная экземпляра этого класса.

    Как видно из следующего примера, Владелец счета создает экземпляр в качестве переменной экземпляра Владелец совместного счета :

    public class AccountHolder {
        private String firstName;
        private String lastName;
        
        AccountHolder jointAccountHolder = new AccountHolder();
    }

    Когда Владелец учетной записи класс создается , a StackOverflowError выбрасывается из-за рекурсивного вызова конструктора, как показано в этом тесте:

    public class AccountHolderManualTest {
        @Test(expected = StackOverflowError.class)
        public void whenInstanciatingAccountHolder_thenThrowsException() {
            AccountHolder holder = new AccountHolder();
        }
    }

    4. Работа С Ошибкой StackOverflowError

    Лучшее, что можно сделать при обнаружении StackOverflowError , – это осторожно проверить трассировку стека, чтобы определить повторяющийся шаблон номеров строк. Это позволит нам найти код, который имеет проблемную рекурсию.

    Лучшее, что можно сделать при обнаружении || StackOverflowError||, – это осторожно проверить трассировку стека, чтобы определить повторяющийся шаблон номеров строк. Это позволит нам найти код, который имеет проблемную рекурсию.

    Эта трассировка стека создается Бесконечной рекурсией С ручным тестом условия завершения , если мы опустим объявление ожидаемого исключения:

    java.lang.StackOverflowError
    
     at c.b.s.InfiniteRecursionWithTerminationCondition
      .calculateFactorial(InfiniteRecursionWithTerminationCondition.java:5)
     at c.b.s.InfiniteRecursionWithTerminationCondition
      .calculateFactorial(InfiniteRecursionWithTerminationCondition.java:5)
     at c.b.s.InfiniteRecursionWithTerminationCondition
      .calculateFactorial(InfiniteRecursionWithTerminationCondition.java:5)
     at c.b.s.InfiniteRecursionWithTerminationCondition
      .calculateFactorial(InfiniteRecursionWithTerminationCondition.java:5)

    Здесь можно увидеть повторение строки № 5. Именно здесь выполняется рекурсивный вызов. Теперь это просто вопрос изучения кода, чтобы увидеть, правильно ли выполняется рекурсия.

    Вот трассировка стека, которую мы получаем, выполняя ручной тест циклической зависимости (опять же, без ожидаемого исключения):

    java.lang.StackOverflowError
      at c.b.s.ClassTwo.(ClassTwo.java:9)
      at c.b.s.ClassOne.(ClassOne.java:9)
      at c.b.s.ClassTwo.(ClassTwo.java:9)
      at c.b.s.ClassOne.(ClassOne.java:9)

    Эта трассировка стека показывает номера строк, которые вызывают проблему в двух классах, находящихся в циклической связи. Строка номер 9 класса Два и строка номер 9 класса Один указывают на местоположение внутри конструктора, где он пытается создать экземпляр другого класса.

    После тщательной проверки кода и если ни одно из следующих действий (или любая другая логическая ошибка кода) не является причиной ошибки:

    • Неправильно реализованная рекурсия (т. е. без условия завершения)
    • Циклическая зависимость между классами
    • Создание экземпляра класса в том же классе, что и переменная экземпляра этого класса

    Было бы неплохо попытаться увеличить размер стека. В зависимости от установленной JVM размер стека по умолчанию может варьироваться.

    Флаг -Xss можно использовать для увеличения размера стека либо из конфигурации проекта, либо из командной строки.

    5. Заключение

    В этой статье мы более подробно рассмотрели StackOverflowError , включая то, как Java-код может вызвать его, а также как мы можем диагностировать и исправить его.

    Исходный код, связанный с этой статьей, можно найти на GitHub .

    The java.lang.stackoverflowerror – StackOverflow Error in Java is thrown to indicate that the application’s stack was exhausted, due to deep recursion.

    The StackOverflowError extends the VirtualMachineError class, which indicates that the JVM is broken, or it has run out of resources and cannot operate. Furthermore, the VirtualMachineError extends the Error class, which is used to indicate those serious problems that an application should not catch. A method may not declare such errors in its throw clause, because these errors are abnormal conditions that shall never occur.

    Finally, the StackOverflowError exists since the 1.0 version of Java.

    You can also check this tutorial in the following video:

    java.lang.StackOverflowError – How to solve StackOverflowError – Video

    1. The Structure of StackOverflowError

    Constructors

    • StackOverflowError()

    Creates an instance of the StackOverflowError class, setting null as its message.

    • StackOverflowError(String s)

    Creates an instance of the StackOverflowError class, using the specified string as message. The string argument indicates the name of the class that threw the error.

    2. The StackOverflowError in Java

    When a function call is invoked by a Java application, a stack frame is allocated on the call stack. The stack frame contains the parameters of the invoked method, its local parameters, and the return address of the method. The return address denotes the execution point from which, the program execution shall continue after the invoked method returns. If there is no space for a new stack frame then, the StackOverflowError is thrown by the Java Virtual Machine (JVM).

    stackoverflow error java - stackoverflowerror

    The most common case that can possibly exhaust a Java application’s stack is recursion. In recursion, a method invokes itself during its execution. Recursion is considered as a powerful general-purpose programming technique, but must be used with caution, in order for the StackOverflowError to be avoided.

    An example that throws a StackOverflowError is shown below:

    StackOverflowErrorExample.java

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    public class StackOverflowErrorExample {

        public static void recursivePrint(int num) {

            System.out.println("Number: " + num);

            if(num == 0)

                return;

            else

                recursivePrint(++num);

        }

        public static void main(String[] args) {

            StackOverflowErrorExample.recursivePrint(1);

        }

    }

    In this example, we define a recursive method, called recursivePrint that prints an integer and then, calls itself, with the next successive integer as an argument. The recursion ends once we invoke the method, passing 0 as a parameter. However, in our example, we start printing numbers from 1 and thus, the recursion will never terminate.

    A sample execution, using the -Xss1M flag that specifies the size of the thread stack to equal to 1MB, is shown below:

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    Number: 1

    Number: 2

    Number: 3

    ...

    Number: 6262

    Number: 6263

    Number: 6264

    Number: 6265

    Number: 6266

    Exception in thread "main" java.lang.StackOverflowError

            at java.io.PrintStream.write(PrintStream.java:480)

            at sun.nio.cs.StreamEncoder.writeBytes(StreamEncoder.java:221)

            at sun.nio.cs.StreamEncoder.implFlushBuffer(StreamEncoder.java:291)

            at sun.nio.cs.StreamEncoder.flushBuffer(StreamEncoder.java:104)

            at java.io.OutputStreamWriter.flushBuffer(OutputStreamWriter.java:185)

            at java.io.PrintStream.write(PrintStream.java:527)

            at java.io.PrintStream.print(PrintStream.java:669)

            at java.io.PrintStream.println(PrintStream.java:806)

            at StackOverflowErrorExample.recursivePrint(StackOverflowErrorExample.java:4)

            at StackOverflowErrorExample.recursivePrint(StackOverflowErrorExample.java:9)

            at StackOverflowErrorExample.recursivePrint(StackOverflowErrorExample.java:9)

            at StackOverflowErrorExample.recursivePrint(StackOverflowErrorExample.java:9)

            ...

    Depending on the JVM’s initial configuration, the results may differ, but eventually the StackOverflowError shall be thrown. This example is a very good example of how recursion can cause problems, if not implemented with caution.

    3. More about the java.lang.stackoverflowerror

    The following example demonstrates the risk of having cyclic relationships between classes:

    StackOverflowErrorToStringExample.java:

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    class A {

        private int aValue;

        private B bInstance = null;

        public A() {

            aValue = 0;

            bInstance = new B();

        }

        @Override

        public String toString() {

            return "";

        }

    }

    class B {

        private int bValue;

        private A aInstance = null;

        public B() {

            bValue = 10;

            aInstance = new A();

        }

        @Override

        public String toString() {

            return "";

        }

    }

    public class StackOverflowErrorToStringExample {

        public static void main(String[] args) {

            A obj = new A();

            System.out.println(obj.toString());

        }

    }

    In this example, we defined two classes, A and B. The class A contains one instance of the B class, while, the B class contains one instance of the A class. Thus, we have a circular dependency between these two classes. Furthermore, each toString method, invokes the corresponding toString method of the other class, and so on, which results in a StackOverflowError.

    A sample execution is shown below:

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    Exception in thread "main" java.lang.StackOverflowError

        at main.java.B.(StackOverflowErrorToStringExample.java:24)

        at main.java.A.(StackOverflowErrorToStringExample.java:9)

        at main.java.B.(StackOverflowErrorToStringExample.java:24)

        at main.java.A.(StackOverflowErrorToStringExample.java:9)

        at main.java.B.(StackOverflowErrorToStringExample.java:24)

        at main.java.A.(StackOverflowErrorToStringExample.java:9)

        ...

    4. How to deal with the java.lang.stackoverflowerror

    • The simplest solution is to carefully inspect the stack trace and detect the repeating pattern of line numbers. These line numbers indicate the code being recursively called. Once you detect these lines, you must carefully inspect your code and understand why the recursion never terminates.
    • If you have verified that the recursion is implemented correctly, you can increase the stack’s size, in order to allow a larger number of invocations. Depending on the Java Virtual Machine (JVM) installed, the default thread stack size may equal to either 512KB, or 1MB. You can increase the thread stack size using the -Xss flag. This flag can be specified either via the project’s configuration, or via the command line. The format of the -Xss argument is:
      -Xss<size>[g|G|m|M|k|K]

    5. Additional knowledge

    • STACKOVERFLOWERROR: CAUSES & SOLUTIONS
    • java.lang.ClassNotFoundException – How to solve Class Not Found Exception
    • Unreachable Statement Java Error – How to resolve it
    • java.lang.NullPointerException Example – How to handle Java Null Pointer Exception (with video)
    • Try Catch Java Example
    • Java Stack Example (with video)
    • Online Java Compiler – What options are there
    • What is null in Java

    6. Download the Eclipse Project

    This was a tutorial about the StackOverflowError in Java.

    Last updated on Oct. 12th, 2021

    Photo of Sotirios-Efstathios Maneas

    Sotirios-Efstathios (Stathis) Maneas is a PhD student at the Department of Computer Science at the University of Toronto. His main interests include distributed systems, storage systems, file systems, and operating systems.

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