Related
I made a test program to test the Runtime.freeMemory() method.
public class RuntimeFreeMemory {
public static void main(String[] args) {
Runtime r = Runtime.getRuntime();
System.out.println(r.freeMemory()); // 246939608
String[] arr = new String[10000000];
for (int i = 0; i < 10000000; i++)
arr[i] = new String();
System.out.println(r.freeMemory()); // 517655928
}
}
Of course, these are big numbers, but I really wanted to test it as numbers like 10000 wouldn't cut it. But when I ran it, I got some unexpected numbers. I thought the end free memory would go down or stay somewhat the same as the initial free memory, but instead they went over double what the initial free memory was. I ran this multiple times and it was always something around these values. Can somebody explain this?
The freeMemory reported may be affected by GC and by the fact that the heap has been expanded because of your array allocations. It might be interesting to also print r.totalMemory at the beginning and at the end and compare.
When the JVM starts, it doesn't actually allocate (by default) whole heap memory upfront - it's just "reserved" and later "committed" if/when needed (at that point, heap is expanded).
You can try to set -Xms and -Xmx explicitly to the same value and will likely get more expected results.
E.g. by using -Xms8g -Xmx8g I got following:
public static void main(String[] args) {
Runtime r = Runtime.getRuntime();
System.out.println(r.freeMemory()); // 8581851136
System.out.println(r.totalMemory()); // 8589934592
System.out.println("Process PID: " + ProcessHandle.current().pid());
String[] arr = new String[100000000];
for (int i = 0; i < 100000000; i++)
arr[i] = new String();
System.out.println(r.freeMemory()); // 5717141504
System.out.println(r.totalMemory()); // 8589934592
}
// compare results when using ONLY -Xmx8g
// 537178112
// 541065216
// 3240407040
// 6104809472
Note: Using -Xms still doesn't (usually) mean that the memory is resident in RAM: Java heap Xms and linux free memory different
I have simple programm:
public class Test {
public static void main(String[] args) {
for (int i = 0; i < 1_000_000; i++) {
System.out.print(1);
}
}
}
And launched profiling. Here are the results:
I assume that memory grows because of this method calls:
public void print(int i) {
write(String.valueOf(i));
}
Is there a way to print int values in the console without memory drawdown?
On local machine I try add if (i % 10000 == 0) System.gc(); to cycle and memory consumption evened out. But the system that checks the solution does not make a decision. I tried to change the values of the step but still does not pass either in memory(should work less than 20mb) or in time(<1sec)
EDIT I try this
String str = "hz";
for (int i = 0; i < 1_000_0000; i++) {
System.out.print(str);
}
But same result:
EDIT2 if I write this code
public class Test {
public static void main(String[] args) {
byte[] bytes = "hz".getBytes();
for (int i = 0; i < 1_000_0000; i++) {
System.out.write(bytes, 0, bytes.length);
}
}
}
I have this
Therefore, I do not believe that Java is making its noises. They would be in both cases.
You need to convert the int into characters without generating a new String each time you do it. This could be done in a couple of ways:
Write a custom "int to characters" method that converts to ASCII bytes in a byte[] (See #AndyTurner's example code). Then write the byte[]. And repeat.
Use ByteBuffer, fill it directly using a custom "int to characters" converter method, and use a Channel to output the bytes when the buffer is full. And repeat.
If done correctly, you should be able to output the numbers without generating any garbage ... other than your once-off buffers.
Note that System.out is a PrintStream wrapping a BufferedOutputStream wrapping a FileOuputStream. And, when you output a String directly or indirectly using one of the print methods, that actually does through a BufferedWriter that is internal to the PrintStream. It is complicated ... and apparently the print(String) method generates garbage somewhere in that complexity.
Concerning your EDIT 1: when you repeatedly print out a constant string, you are still apparently generating garbage. I was surprised by this, but I guess it is happening in the BufferedWriter.
Concerning your EDIT 2: when you repeatedly write from a byte[], the garbage generation all but disappears. This confirms that at least one of my suggestions will work.
However, since you are monitoring the JVM with an external profile, your JVM is also running an agent that is periodically sending updates to your profiler. That agent will most likely be generating a small amount of garbage. And there could be other sources of garbage in the JVM; e.g. if you have JVM GC logging enabled.
Since you have discovered that printing a byte[] keeps memory allocation within the required bounds, you can use this fact:
Allocate a byte array the length of the ASCII representation of Integer.MIN_VALUE (11 - the longest an int can be). Then you can fill the array backwards to convert a number i:
int p = buffer.length;
if (i == Integer.MIN_VALUE) {
buffer[--p] = (byte) ('0' - i % 10);
i /= 10;
}
boolean neg = i < 0;
if (neg) i = -i;
do {
buffer[--p] = (byte) ('0' + i % 10);
i /= 10;
} while (i != 0);
if (neg) buffer[--p] = '-';
Then write this to your stream:
out.write(buffer, p, buffer.length - p);
You can reuse the same buffer to write as many numbers as you wish.
The pattern of memory usage is typical for java. Your code is irrelevant. To control java memory usage you need to use some -X parameters for example "-Xms512m -Xmx512m" will set both minimum and maximum heap size to 512m. BTW in order to minimize the sow-like memory graph it would be recommended to set min and max size to the same value. Those params could be given to java on command line when you run your java for example:
java -Xms512m -Xmx512m myProgram
There are other ways as well. Here is one link where you can read more about it: Oracle docs. There are other params that control stacksize and some other things. The code itself if written without memory usage considerations may influence memory usage as well, but in your case its too trivial of a code to do anything. Most memory concerns are addressed by configuring jvm memory usage params
What is a StackOverflowError, what causes it, and how should I deal with them?
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).
To describe this, first let us understand how local variables and objects are stored.
Local variable are stored on the stack:
If you looked at the image you should be able to understand how things are working.
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).
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 it must be used with caution, to avoid StackOverflowError.
An example of throwing a StackOverflowError is shown below:
StackOverflowErrorExample.java:
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 until we pass in 0 as a parameter. However, in our example, we passed in the parameter from 1 and its increasing followers, consequently, the recursion will never terminate.
A sample execution, using the -Xss1M flag that specifies the size of the thread stack to equal to 1 MB, is shown below:
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.
How to deal with the 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 512 KB, or 1 MB. 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]
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.
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.
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.
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).
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.
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).
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.
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);
}
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:
Save the return address when making a procedure call
Save local variables
Save special registers to restore them later
Pass arguments to a procedure call (more than 6)
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.
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]);
}
}
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();
}
}
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.
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.
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.
I've run into a really strange bug, and I'm hoping someone here can shed some light as it's way out of my area of expertise.
First, relevant background information: I am running OS X 10.9.4 on a Late 2013 Macbook Pro Retina with a 2.4GHz Haswell CPU. I'm using JDK SE 8u5 for OS X from Oracle, and I'm running my code on the latest version of IntelliJ IDEA. This bug also seems to be specific only to OS X, as I posted on Reddit about this bug already and other users with OS X were able to recreate it while users on Windows and Linux, including myself, had the program run as expected with the println() version running half a second slower than the version without println().
Now for the bug: In my code, I have a println() statement that when included, the program runs at ~2.5 seconds. If I remove the println() statement either by deleting it or commenting it out, the program counterintuitively takes longer to run at ~9 seconds. It's extremely strange as I/O should theoretically slow the program down, not make it faster.
For my actual code, it's my implementation of Project Euler Problem 14. Please keep in mind I'm still a student so it's not the best implementation:
public class ProjectEuler14
{
public static void main(String[] args)
{
final double TIME_START = System.currentTimeMillis();
Collatz c = new Collatz();
int highestNumOfTerms = 0;
int currentNumOfTerms = 0;
int highestValue = 0; //Value which produces most number of Collatz terms
for (double i = 1.; i <= 1000000.; i++)
{
currentNumOfTerms = c.startCollatz(i);
if (currentNumOfTerms > highestNumOfTerms)
{
highestNumOfTerms = currentNumOfTerms;
highestValue = (int)(i);
System.out.println("New term: " + highestValue); //THIS IS THE OFFENDING LINE OF CODE
}
}
final double TIME_STOP = System.currentTimeMillis();
System.out.println("Highest term: " + highestValue + " with " + highestNumOfTerms + " number of terms");
System.out.println("Completed in " + ((TIME_STOP - TIME_START)/1000) + " s");
}
}
public class Collatz
{
private static int numOfTerms = 0;
private boolean isFirstRun = false;
public int startCollatz(double n)
{
isFirstRun = true;
runCollatz(n);
return numOfTerms;
}
private void runCollatz(double n)
{
if (isFirstRun)
{
numOfTerms = 0;
isFirstRun = false;
}
if (n == 1)
{
//Reached last term, does nothing and causes program to return to startCollatz()
}
else if (n % 2 == 0)
{
//Divides n by 2 following Collatz rule, running recursion
numOfTerms = numOfTerms + 1;
runCollatz(n / 2);
}
else if (n % 2 == 1)
{
//Multiples n by 3 and adds one, following Collatz rule, running recursion
numOfTerms = numOfTerms + 1;
runCollatz((3 * n) + 1);
}
}
}
The line of code in question has been commented in with all caps, as it doesn't look like SO does line numbers. If you can't find it, it's within the nested if() statement in my for() loop in my main method.
I've run my code multiple times with and without that line, and I consistently get the above stated ~2.5sec times with println() and ~9sec without println(). I've also rebooted my laptop multiple times to make sure it wasn't my current OS run and the times stay consistent.
Since other OS X 10.9.4 users were able to replicate the code, I suspect it's due to a low-level bug with the compliler, JVM, or OS itself. In any case, this is way outside my knowledge. It's not a critical bug, but I definitely am interested in why this is happening and would appreciate any insight.
I did some research and some more with #ekabanov and here are the findings.
The effect you are seeing only happens with Java 8 and not with Java 7.
The extra line triggers a different JIT compilation/optimisation
The assembly code of the faster version is ~3 times larger and quick glance shows it did loop unrolling
The JIT compilation log shows that the slower version successfully inlined the runCollatz while the faster didn't stating that the callee is too large (probably because of the unrolling).
There is a great tool that helps you analyse such situations, it is called jitwatch. If it is assembly level then you also need the HotSpot Disassembler.
I'll post also my log files. You can feed the hotspot log files to the jitwatch and the assembly extraction is something that you diff to spot the differences.
Fast version's hotspot log file
Fast version's assembly log file
Slow version's hotspot log file
Slow version's assembly log file
What is a StackOverflowError, what causes it, and how should I deal with them?
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).
To describe this, first let us understand how local variables and objects are stored.
Local variable are stored on the stack:
If you looked at the image you should be able to understand how things are working.
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).
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 it must be used with caution, to avoid StackOverflowError.
An example of throwing a StackOverflowError is shown below:
StackOverflowErrorExample.java:
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 until we pass in 0 as a parameter. However, in our example, we passed in the parameter from 1 and its increasing followers, consequently, the recursion will never terminate.
A sample execution, using the -Xss1M flag that specifies the size of the thread stack to equal to 1 MB, is shown below:
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.
How to deal with the 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 512 KB, or 1 MB. 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]
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.
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.
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.
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).
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.
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).
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.
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);
}
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:
Save the return address when making a procedure call
Save local variables
Save special registers to restore them later
Pass arguments to a procedure call (more than 6)
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.
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]);
}
}
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();
}
}
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.
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.
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.