I have a favorite C# program similar to the one below that shows that if two threads share the same memory address for counting (one thread incrementing n times, one thread decrementing n times) you can get a final result other than zero. As long as n is reasonably large, it's pretty easy to get C# to display some non-zero value between [-n, n]. However, I can't get Java to produce a non-zero result even when increasing the number of threads to 1000 (500 up, 500 down). Is there some memory model or specification difference wrt C# I'm not aware of that guarantees this program will always yield 0 despite the scheduling or number of cores that I am not aware of? Would we agree that this program could produce a non-zero value even if we can not prove that experimentally?
(Not:, I found this exact question over here, but when I run that topic's code I also get zero.)
public class Counter
{
private int _counter = 0;
Counter() throws Exception
{
final int limit = Integer.MAX_VALUE;
Thread add = new Thread()
{
public void run()
{
for(int i = 0; i<limit; i++)
{
_counter++;
}
}
};
Thread sub = new Thread()
{
public void run()
{
for(int i = 0; i<limit; i++)
{
_counter--;
}
}
};
add.run();
sub.run();
add.join();
sub.join();
System.out.println(_counter);
}
public static void main(String[] args) throws Exception
{
new Counter();
}
}
The code you've given only runs on a single thread, so will always give a result of 0. If you actually start two threads, you can indeed get a non-zero result:
// Don't call run(), which is a synchronous call, which doesn't start any threads
// Call start(), which starts a new thread and calls run() *in that thread*.
add.start();
sub.start();
On my box in a test run that gave -2146200243.
Assuming you really meant start, not run.
On most common platforms it will very likely produce non zero, because ++/-- are not atomic operations in case of multiple cores. On single core/single CPU you will most likely get 0 because ++/-- are atomic if compiled to one instruction (add/inc) but that part depends on JVM.
Check result here: http://ideone.com/IzTT2
The problem with your program is that you are not creating an OS thread, so your program is essentially single threaded. In Java you must call Thread.start() to create a new OS thread, not Thread.run(). This has to do with a regrettable mistake made in the initial Java API. That mistake is that the designer made Thread implement Runnable.
add.start();
sub.start();
add.join();
sub.join();
Related
I tried to compile the example from Thinking in Java by Bruce Eckel:
import java.util.concurrent.*;
public class SimplePriorities implements Runnable {
private int countDown = 5;
private volatile double d; // No optimization
private int priority;
public SimplePriorities(int priority) {
this.priority = priority;
}
public String toString() {
return Thread.currentThread() + ": " + countDown;
}
public void run() {
Thread.currentThread().setPriority(priority);
while(true) {
// An expensive, interruptable operation:
for(int i = 1; i < 100000; i++) {
d += (Math.PI + Math.E) / (double)i;
if(i % 1000 == 0)
Thread.yield();
}
System.out.println(this);
if(--countDown == 0) return;
}
}
public static void main(String[] args) {
ExecutorService exec = Executors.newCachedThreadPool();
for(int i = 0; i < 5; i++)
exec.execute(
new SimplePriorities(Thread.MIN_PRIORITY));
exec.execute(
new SimplePriorities(Thread.MAX_PRIORITY));
exec.shutdown();
}
}
According to the book, the output has to look like:
Thread[pool-1-thread-6,10,main]: 5
Thread[pool-1-thread-6,10,main]: 4
Thread[pool-1-thread-6,10,main]: 3
Thread[pool-1-thread-6,10,main]: 2
Thread[pool-1-thread-6,10,main]: 1
Thread[pool-1-thread-3,1,main]: 5
Thread[pool-1-thread-2,1,main]: 5
Thread[pool-1-thread-1,1,main]: 5
...
But in my case 6th thread doesn't execute its task at first and threads are disordered. Could you please explain me what's wrong? I just copied the source and didn't add any strings of code.
The code is working fine and with the output from the book.
Your IDE probably has console window with the scroll bar - just scroll it up and see the 6th thread first doing its job.
However, the results may differ depending on OS / JVM version. This code runs as expected for me on Windows 10 / JVM 8
There are two issues here:
If two threads with the same priority want to write output, which one goes first?
The order of threads (with the same priority) is undefined, therefore the order of output is undefined. It is likely that a single thread is allowed to write several outputs in a row (because that's how most thread schedulers work), but it could also be completely random, or anything in between.
How many threads will a cached thread pool create?
That depends on your system. If you run on a dual-core system, creating more than 4 threads is pointless, because there hardly won't be any CPU available to execute those threads. In this scenario further tasks will be queued and executed only after earlier tasks are completed.
Hint: there is also a fixed-size thread pool, experimenting with that should change the output.
In summary there is nothing wrong with your code, it is just wrong to assume that threads are executed in any order. It is even technically possible (although very unlikely), that the first task is already completed before the last task is even started. If your book says that the above order is "correct" then the book is simply wrong. On an average system that might be the most likely output, but - as above - with threads there is never any order, unless you enforce it.
One way to enforce it are thread priorities - higher priorities will get their work done first - you can find other concepts in the concurrent package.
I am running the below class .
public class RunThreads implements Runnable {
static int i;
public static void main(String[] args) {
RunThreads job = new RunThreads();
Thread alpha = new Thread(job);
Thread beta = new Thread(job);
alpha.setName("Alpha");
beta.setName("beta");
alpha.start();
beta.start();
}
public void run(){
for(;i<10;i++){
System.out.println(Thread.currentThread().getName() + i);
}
}
}
And my output is :
beta0
beta1
Alpha0
beta2
beta4
beta5
beta6
Alpha3
Alpha8
beta7
Alpha9
I understand I will get different outputs every time I execute it. My question is, why does the output have the value of i as 0 twice, for both the alpha and beta threads i.e. Alpha0 and beta0.
The value of i has been incremented to 1 by the beta thread. So, how does the alpha thread print out Alpha0
I maybe missing something very obvious here. Thanks !
Things are scary when you access shared data with no synchronization etc:
There's no guarantee that the Alpha thread reading i will see the "latest" update from the beta thread or vice versa
It's possible for both threads to start i++ at roughly the same time... and as i++ is basically:
int tmp = i;
tmp++;
i = tmp;
you can easily "lose" increments
If you want to make this thread-safe, you should use AtomicInteger instead:
import java.util.concurrent.atomic.AtomicInteger;
public class RunThreads implements Runnable {
static AtomicInteger counter = new AtomicInteger();
public static void main(String[] args) {
RunThreads job = new RunThreads();
Thread alpha = new Thread(job);
Thread beta = new Thread(job);
alpha.setName("Alpha");
beta.setName("beta");
alpha.start();
beta.start();
}
public void run(){
int local;
while ((local = counter.getAndIncrement()) < 10) {
System.out.println(Thread.currentThread().getName() + local);
}
}
}
Your output may still appear to be in the wrong order (because the alpha thread may "start" writing "Alpha0" while the beta thread "starts" writing "beta1" but the beta thread gets the lock on console output first), but you'll only see each count once. Note that you have to use the result of getAndIncrement() for both the checking and the printing - if you called counter.get() in the body of the loop, you could still see duplicates due to the interleaving of operations.
The simple answer here would be that whether or not the variable is volatile or atomic or whatever, both your threads start out when the variable value is 0, and only change it after the print.
This means that both threads can reach the "print" line before either of them reached the i++.
Which means that both are very likely to print 0, unless one of them is delayed long enough for the other one to update the variable (at which point the question of memory model and data visibility arise).
"While" thread A is doing its update, thread B is reading that value.
Like in:
A: prints i (current value 0)
B: prints i (current value 0)
A: stores 1 to i
You can't assume that those two actions:
for(;i<10;i++){ // increasing the loop counter
and
System.out.println(Thread.currentThread().getName() + i);
printing the loop counter happen in "one shot". To the contrary.
Both threads execute the corresponding instructions in parallel; and there are absolutely no guarantees on the outcome of that.
Threads might run on separate cores in multicore architecture and each core has it's own registry set or local cache, as long as you don't invalidate a cache the running core might not reach to a RAM memory to get up-to-date value and will use the cached one. To make it work as you expected it the variable would have to be marked as volatile, which would invalidate a cache on each write to this memory location and fetch the value directly from RAM.
Update:
As was pointed in comments - volatile is not enough to keep the value up to date, there is a read and write operation on i++. To make it work AtomicInteger would be enough, or slightly more expensive locking around i++.
I'm trying to convert this code to java and using thread to implement it
turn = 0 // shared control variable
while (turn != i);
// CS
turn = (turn + 1) % n;
I'm really tried hard to reach to right code but I failed this is my code
/*
* Mutual exclusion using thread
*/
class gV{
int turn=0;
}
class newThread extends Thread{
static int i;
int n=10;
newThread(gV obj){
this.i=obj.turn;
start();
}
public void run(){
while(obj.turn!=i&&obj.turn<n);
criticalSection(i);
obj.turn=(obj.turn+1);
i++;
}
public void criticalSection(int numOfProcess){
System.out.println("Process " + numOfProcess + " done!!");
}
}
class MutualExclusion{
public static void main(String args[]){
gV obj = new gV();
new newThread(obj);
}
}
I know my code has some mistakes. Thank you for the help!
Use an AtomicInteger.
Atomic means that any operation on it will fully complete before any other thread can see the result. Meaning that you won't have two simultaneous operations 'clobber' it. For example, imagine if you had a non atomic integer and two threads attempted to increment it simultaneously - say it had value 1, they both read it as 1 and attempt to set it to 2. They both incremented it once - but instead of it becoming 3, it became 2! AtomicInteger solves this problem by giving you IncrementAndGet, which guarantees no other thread can access the AtomicInteger's value before the increment completes.
In particular, use these methods:
http://docs.oracle.com/javase/6/docs/api/java/util/concurrent/atomic/AtomicInteger.html#get()
http://docs.oracle.com/javase/6/docs/api/java/util/concurrent/atomic/AtomicInteger.html#incrementAndGet()
You might notice that this increments it, but it doesn't take it modulo n. Well, you can take it modulo n whenever you read its value, you don't need it to be stored that way.
EDIT: By the way, doing something like this:
while (turn != i);
is called busy-waiting, and it's a bad idea because it means that CPU usage will be 100%, checking the variable hundreds of thousands of times per second. In this kind of scenario, instead of making each thread check as often as possible, you want to have threads wait and be notifyed by another thread when it is that thread's turn to continue execution.
I believe in Java that using lock and synchronized to implement mutual exclusion will also give you this property, e.g. if you try to lock on something or enter a synchronized block but it is already in use then the thread goes to sleep and is woken up when it is its turn. So, you can look into this as well.
I'm attempting to test some benchmarking tools by running them against a simple program which increments a variable as many times as possible for 1000 milliseconds.
How many incrementations of a single 64 bit number should I expect to be able to perform on an intel i7 chip on the JDK for Mac OS X ?
My current methodology is :
start thread (t2) that continually increments "i" in an infinite loop (for(;;;)).
let the main thread (call it t1) sleep for 1000 milliseconds.
have t1 interrupt (or stop, since this deprecated method works on Apple's JDK 6) t2.
Currently, I am reproducibly getting about 2E8 incrementations (this is tabulated below: the value shown is the value that is printed when the incrementing thread is interrupted after a 1000 millisecond sleep() in the calling thread).
217057470
223302277
212337757
215177075
214785738
213849329
215645992
215651712
215363726
216135710
How can I know wether this benchmark is reasonable or not, i.e., what is the theoretical fastest speed at which an i7 chip should be able to increment a single 64-bit digit? This code is running in the JVM and is below:
package net.rudolfcode.jvm;
/**
* How many instructions can the JVM exeucte in a second?
* #author jayunit100
*/
public class Example3B {
public static void main(String[] args){
for(int i =0 ; i < 10 ; i++){
Thread addThread = createThread();
runForASecond(addThread,1000);
}
}
private static Thread createThread() {
Thread addThread = new Thread(){
Long i =0L;
public void run() {
boolean t=true;
for (;;) {
try {
i++;
}
catch (Exception e) {
e.printStackTrace();
}
}
}
#Override
public void interrupt() {
System.out.println(i);
super.interrupt();
}
};
return addThread;
}
private static void runForASecond(Thread addThread, int milli) {
addThread.start();
try{
Thread.sleep(milli);
}
catch(Exception e){
}
addThread.interrupt();
//stop() works on some JVMs...
addThread.stop();
}
}
Theoretically, making some assumptions which are probably not valid:
Assume that a number can be incremented in 1 instruction (probably not, because you're running in a JVM and not natively)
Assume that a 2.5 GHz processor can execute 2,500,000,000 instructions per second (but in reality, it's more complicated than that)
Then you could say that 2,500,000,000 increments in 1 second is a "reasonable" upper bound based on the simplest possible back-of-the-envelope estimation.
How far off is that from your measurement?
2,500,000,000 is O(1,000,000,000)
2E8 is O(100,000,000)
So we're only off by 1 order of magnitude. Given the wildly unfounded assumptions – sounds reasonable to me.
First of all beware of JVM optimisations! You must be sure you measure exactly what you think you do. Since Long i =0L; is not volatile and it's effectively useless (nothing is done to intermediate values) JIT can do pretty nasty stuff.
As for the estimation you can think of not more then X*10^9 operations per second on X GHz machine. You can safely divide this value for 10 for probably because instructions aren't mapped 1:1.
So you're pretty close :)
I'm trying to understand synchronization of multiple threads in Java more fully. I understand the high level idea behind the use of the synchronized keyword, and how it provides mutual exclusion among threads.
The only thing is that most of the examples I read online and in my textbook still work correctly even if you remove the synchronized keyword which is making this topic more confusing than I think it needs to be.
Can anyone provide me with a concrete example of when not including the synchronized keyword will produce erroneous results? Any information would be much appreciated.
You can usually trigger a race condition by increasing the number of iterations. Here's a simple example that works with 100 and 1,000 iterations but fails (at least on my quad-core box) at 10,000 iterations (sometimes).
public class Race
{
static final int ITERATIONS = 10000;
static int counter;
public static void main(String[] args) throws InterruptedException {
System.out.println("start");
Thread first = new Thread(new Runnable() {
#Override
public void run() {
for (int i = 0; i < ITERATIONS; i++) {
counter++;
}
}
});
Thread second = new Thread(new Runnable() {
#Override
public void run() {
for (int i = 0; i < ITERATIONS; i++) {
counter++;
}
}
});
first.start();
second.start();
first.join();
second.join();
System.out.println("Counter " + counter + " should be " + (2 * ITERATIONS));
}
}
>>> Counter 12325 should be 20000
This example fails because access to counter is not properly synchronized. It can fail in two ways, possibly both in the same run:
One thread fails to see that the other has incremented the counter because it doesn't see the new value.
One thread increments the counter between the other thread reading the current value and writing the new value. This is because the increment and decrement operators are not atomic.
The fix for this simple program would be to use an AtomicInteger. Using volatile isn't enough due to the problem with increment, but AtomicInteger provides atomic operations for increment, get-and-set, etc.
The thing about race conditions is that they don't necessarily happen if you don't do proper synchronization -- indeed, quite frequently it'll work just fine -- but then one year later, in the middle of the night, your code will crash with a completely unpredictable bug that you can't reproduce, because the bug only appears randomly.
Race conditions are so insidious precisely because they don't always make your program crash, and they trigger more or less randomly.