I am confused at the output of the following program
public class ChicksYack implements Runnable {
Chicks c ;
public static void main(String[] args){
new ChicksYack().go();
}
void go(){
c= new Chicks();
new Thread(new ChicksYack()).start();
new Thread(new ChicksYack()).start();
}
public void run() {
c.yack(Thread.currentThread().getId());
}
}
class Chicks{
synchronized void yack(long id){
for(int x = 1 ; x < 3 ; x++){
System.out.print(id + " ");
Thread.yield();
}
}
}
The program throws NullPointerException at run-time. Will the value of Chicks variable c won't be shared on thread-1 and thread-2 stack. I know i am making a pretty silly mistake but pretty confused. Any pointers will be helpful.
If you're ensuring that there is an instance of Chicks in each instance of ChicksYack, then you need to move the line:
c= new Chicks();
into a constructor e.g.
public ChicksYack() {
c= new Chicks();
}
and change the type of the member variable from Chicks c to final Chicks c; otherwise you are not guaranteed that you will see it fully constructed in the threads. Adding the final keywoard ensures that:
When the constructor exits, the values of final fields are guaranteed to be visible to other threads accessing the constructed object. (ref: Javamex.com website)
As is, you're creating two more instances of ChicksYack that don't have the c member initialized.
Now if you're intending on having only a single instance of Chicks shared amongst all the instances of ChicksYack, then you need to declare it as static e.g.
static Chicks c;
you have three instances of ChicksYack and not-static field for Chicks.
I think you should change go method to something like this
void go(){
c= new Chicks();
new Thread(this).start();
new Thread(this).start();
}
The program throws NullPointerException at run-time. Will the value of Chicks variable c won't be shared on thread-1 and thread-2 stack.
Nothing in a computer really happen immediately. If you believed anything was instant, forget that. This means that while a thread can update a field, it takes time to be visible to other threads.
There is a number of exceptions, one is using volatile fields (It is still not instant instead it waits for any update) and the other is any field set before a thread is started. If this program really throws an NPE for you I would suspect you have a bug in your JVM. I would make sure you have the latest version
Related
So I am trying to pass arguments to my threads and my output is not as expected. Below is my code:
class BadThreadParam implements Runnable {
static int c;
public BadThreadParam( int a, int b ) {
c = a + b;
}
public void run() {
System.out.println( c );
}
}
public class BadThreadParamTest {
public static void main( String[] args ) {
BadThreadParam shouldBe3 = new BadThreadParam( 1, 2 );
BadThreadParam shouldBe5 = new BadThreadParam( 3, 12 );
shouldBe3.run(); // Expect 3 but is 15. WTF?
shouldBe5.run(); // Expect 15.
}
}
The final output I am expecting is:
3
15
but I am getting:
15
15
So is this a problem of thread interference? Can anyone explain why this is happening?
What if you change
static int c;
to
private int c;
Since you don't access the variable c out of your class, there is no reason it should be static. If you make it static, each time you change it, it will be that last value.
If you change it to private it's an 'instance' variable that is linked to the instance you create with new BadThreadParam(); so the output is the value that is given to that specific instance.
c is a static variable. This means that when one thread changes it, it changes for everyone.
You probably intend for it to be an instance variable.
I just re-ordered the lines in your main method and here is the difference
public static void main(String[] args) {
BadThreadParam shouldBe3 = new BadThreadParam(1, 2);
shouldBe3.run();
BadThreadParam shouldBe5 = new BadThreadParam(3, 12);
shouldBe5.run();
}
And here is the output
3
15
Have a look at this along with the answers posted by #JC97 & #Joe C
Hope this will clear your doubts about the static variables
You asked if the problem you're seeing is caused by thread interference. There is no possibility for thread interference in the posted program, because there's only one thread.
You're calling run on a Runnable, which executes it in the current thread.
If you want to create a new thread to execute each of your Runnables then you should change the code to
new Thread(shouldBe3).start();
new Thread(shouldBe15).start();
and then you will have a program with multiple threads. At that point there will be no guarantee that one thread will print its output before the other.
To confirm this you can add the line
System.out.println(Thread.currentThread().getName());
to the body of each run method and to the main method. If they all print the same name then you aren't creating any threads.
As the other answers say, you're letting the second constructor call overwrite the contents of c set by the first constructor call. The static keyword means the variable belongs to the class, not to any one instance, and all instances access the same variable. c should be an instance variable (delete the static keyword) so that each Runnable object has its own copy.
Is it bad programming to initialize two threads with the same instance of a runnable? What difference would it make to initialize with separate instances of a runnable, and does sharing memory locations at all for the same instance of a runnable have anything to do with performance?
public static void main(String[] args)throws Exception {
H h = new H();
H h2 = new H();
Thread j = new Thread(h);
j.setName("11");
Thread jj = new Thread(h);//instead of new H()
jj.setName("22");
j.start();
jj.start();
}
class H implements Runnable {
public void run() {
while(true) {
System.out.println(Thread.currentThread().getName());
}
}
}
It's absolutely fine to do it so long as the code you're running is designed to support that. Not only will it save some memory by having a single instance instead of multiple instances, but if those threads are trying to communicate via shared data, then it may be absolutely required!
Admittedly communicating via shared state is where threading often gets tricky, so this needs to be done carefully, but from the point of view of the threading system itself, there's absolutely no problem in having two threads call the run method of a single Runnable instance.
Since H doesn't have any instance state, using multiple instances won't matter. You need to take care when the Runnable instances start storing state.
public class Main implements Runnable {
volatile int i;
public void run() {
for (i = 0; i < 100; i++) {
System.out.println(i);
}
}
public static void main(String[] args) {
Main a = new Main();
Thread t1 = new Thread(a);
Thread t2 = new Thread(a);
t1.start();
t2.start();
}
}
What gets printed? When you do need to share state between threads, it's a good idea to use the classes in java.util.concurrent. They were written primarily by an expert in multithreading (Doug Lea, author of Concurrent Programming in Java) and tested by many people. Save yourself some heartache. :)
Is it bad programming to initialize two threads with the same instance of a runnable?
Not specifically. However, if the Runnable instance has instance fields, then you'll need to make sure that all access to the fields by the thread is properly synchronized, and this will make the code more complicated.
What difference would it make to initialize with separate instances of a runnable, and does sharing memory locations at all for the same instance of a runnable have anything to do with performance?
The memory saved by sharing a Runnable instance between multiple threads is insignificant ... unless the Runnable holds a significant amount of instance data. (And if it does, the chances are that this will make the instance non-shareable.)
Your H class is an example where sharing instances is safe, but pointless since the memory saving is insignificant. (A Runnable object with no instance fields occupies roughly 8 to 16 bytes, depending on the platform.)
To make understand easily(based on the comment of Stephen), added the below program block about the impact of accessing the instance variable from a non-synchronized block with the same instance of Runnable displays the unexpected results.
public class SynchronizedInstanceMethod implements Runnable{
private int counter;
public SynchronizedInstanceMethod(int counterValue){
this.counter = counterValue;
}
private synchronized void displayMessage(){
System.out.println(" Display Message ");
}
private void modifyCounter(){
this.counter++;
System.out.println("Value -- "+ this.counter);
}
#Override
public void run() {
this.displayMessage();
this.modifyCounter();
}
public static void main(String[] args) {
SynchronizedInstanceMethod instance = new SynchronizedInstanceMethod(5);
new Thread(instance).start();
new Thread(instance).start();
}
}
im trying to write a program in which two threads are created and the output should be like 1st thread prints 1 and the next thread prints 2 ,1st thread again prints 3 and so on. im a beginner so pls help me clearly. i thought thread share the same memory so they will share the i variable and print accordingly. but in output i get like thread1: 1, thread2 : 1, thread1: 2, thread2 : 2 nd so on. pls help. here is my code
class me extends Thread
{
public int name,i;
public void run()
{
for(i=1;i<=50;i++)
{
System.out.println("Thread" + name + " : " + i);
try
{
sleep(1000);
}
catch(Exception e)
{
System.out.println("some problem");
}
}
}
}
public class he
{
public static void main(String[] args)
{
me a=new me();
me b=new me();
a.name=1;
b.name=2;
a.start();
b.start();
}
}
First off you should read this http://www.oracle.com/technetwork/java/codeconventions-135099.html.
Secondly the class member variables are not shared memory. You need to explicitly pass an object (such as the counter) to both objects, such that it becomes shared. However, this will still not be enough. The shared memory can be cached by the threads so you will have race-conditions. To solve this you will need to use a Lock or use an AtomicInteger
It seems what you want to do is:
Write all numbers from 1 to 50 to System.out
without any number being printed multiple times
with the numbers being printed in order
Have this execution be done by two concurrent threads
First, let's look at what is happening in your code: Each number is printed twice. The reason for this is that i is an instance variable of me, your Thread. So each Thread has its own i, i.e., they do not share the value.
To make the two threads share the same value, we need to pass the same value when constructing me. Now, doing so with the primitive int won't help us much, because by passing an int we are not passing a reference, hence the two threads will still work on independent memory locations.
Let us define a new class, Value which holds the integer for us: (Edit: The same could also be achieved by passing an array int[], which also holds the reference to the memory location of its content)
class Value{
int i = 1;
}
Now, main can instantiate one object of type Value and pass the reference to it to both threads. This way, they can access the same memory location.
class Me extends Thread {
final Value v;
public Me(Value v){
this.v = v;
}
public void run(){
for(; v.i < 50; v.i++){
// ...
}
public static void main(){
Value valueInstance = new Value();
Me a = new Me(valueInstance);
Me b = new Me(valueInstance);
}
}
Now i isn't printed twice each time. However, you'll notice that the behavior is still not as desired. This is because the operations are interleaved: a may read i, let's say, the value is 5. Next, b increments the value of i, and stores the new value. i is now 6. However, a did still read the old value, 5, and will print 5 again, even though b just printed 5.
To solve this, we must lock the instance v, i.e., the object of type Value. Java provides the keyword synchronized, which will hold a lock during the execution of all code inside the synchronized block. However, if you simply put synchronize in your method, you still won't get what you desire. Assuming you write:
public void run(){ synchronized(v) {
for(; v.i < 50; v.i++) {
// ...
}}
Your first thread will acquire the lock, but never release it until the entire loop has been executed (which is when i has the value 50). Hence, you must release the lock somehow when it is safe to do so. Well... the only code in your run method that does not depend on i (and hence does not need to be locking) is sleep, which luckily also is where the thread spends the most time in.
Since everything is in the loop body, a simple synchronized block won't do. We can use Semaphore to acquire a lock. So, we create a Semaphore instance in the main method, and, similar to v, pass it to both threads. We can then acquire and release the lock on the Semaphore to let both threads have the chance to get the resource, while guaranteeing safety.
Here's the code that will do the trick:
public class Me extends Thread {
public int name;
final Value v;
final Semaphore lock;
public Me(Value v, Semaphore lock) {
this.v = v;
this.lock = lock;
}
public void run() {
try {
lock.acquire();
while (v.i <= 50) {
System.out.println("Thread" + name + " : " + v.i);
v.i++;
lock.release();
sleep(100);
lock.acquire();
}
lock.release();
} catch (Exception e) {
System.out.println("some problem");
}
}
public static void main(String[] args) {
Value v = new Value();
Semaphore lock = new Semaphore(1);
Me a = new Me(v, lock);
Me b = new Me(v, lock);
a.name = 1;
b.name = 2;
a.start();
b.start();
}
static class Value {
int i = 1;
}
}
Note: Since we are acquiring the lock at the end of the loop, we must also release it after the loop, or the resource will never be freed. Also, I changed the for-loop to a while loop, because we need to update i before releasing the lock for the first time, or the other thread can again read the same value.
Check the below link for the solution. Using multiple threads we can print the numbers in ascending order
http://cooltekhie.blogspot.in/2017/06/#987628206008590221
What I understand by synchronizing static object which is a variable, if one thread is accessing it, other thread can't.
class T3
{
static Integer i = 0;
static void callStatic()
{
synchronized(T3.class)
{
System.out.println(i++);
while(true);
}
}
public void notStatic()
{
System.out.println(i++);
while(true);
}
}
class T2 implements Runnable
{
public void run()
{
System.out.println("calling nonstatic");
new T3().notStatic();
}
}
class T implements Runnable
{
public void run()
{
System.out.println("calling static");
T3.callStatic();
}
}
public class Test
{
public static void main(String[] args)
{
new Thread(new T()).start();
try
{
Thread.sleep(100);
}
catch (InterruptedException e)
{
}
new Thread(new T2()).start();
}
}
But this demo program has output as :
calling static
0
calling nonstatic
1
Is my understanding wrong? Or am I missing something?
I tried, synchronzing callStatic method, and synchronizing T3.class class object too. But none worked as I thought.
Note : I thought, 1 will not be printed as callStatic has lock on variable i and is in infinite loop.
You don't synchronize on variables, you synchronize on objects.
callStatic synchronizes on 1 and then sets i to 2. If notStatic were to enter a synchronized(i) block at this point, it would synchronize on 2. No other thread has locked 2, so it proceeds.
(Actually, 1 and 2 aren't objects, but Integer.valueOf(1) and Integer.valueOf(2) return objects, and the compiler automatically inserts the Integer.valueOf calls to convert ints to Integers)
In your code, notStatic doesn't actually synchronize at all. It's true that only one thread can be in a synchronized block for a particular object at a particular time, but that has no effect on other threads that are not trying to enter a synchronized block.
Note: This answer relates to the original question, which had synchronized(i), not synchronized(T3.class), in callStatic. The edit really changes the question.
synchronize acts on the object, not the variable/member holding it. There are a couple of important things going on in your code.
synchronize(i) does indeed synchronize access to i provided that the other code trying to use it also synchronizes. It has no effect on code that doesn't synchronize. Suppose Thread A does synchronize(i) and holds it (your infinite loop); then Thread B does System.out.println(i); Thread B can happily read i, there's nothing stopping it. Thread B would have to do
synchronize (i) {
System.out.println(i);
}
...in order to be affected by Thread A's synchronize(i). Your code is (attempting to) synchronized mutation, but not access.
i++; with an Integer is effectively equivalent to i = new Integer(i.intValue() + 1), because Integer is immutable. So it creates a different Integer object and stores that in the i member. So anything synchronizing on the old Integer object has no effect on code synchronizing on the new one. So even if your code were synchronizing both access and mutation, it wouldn't matter, because the synch would be on the old object.
This means that the code in your callStatic is synchronizing on an instance of Integer and then repeated creating a bunch of other instances, which it is not synchronizing on.
Synchronized blocks on static and non static don't block each other. You need to understand how synchronized works for this. Synchronized is done always on an object never on a variable. When you synchronize, thread takes a lock on the object's monitor, the object which you put in the synchronized statement braces.
Synchronized blocks on static references (like in your code) lock on the .class object of your class and not on any instance of that class. So static and non-static synchronized blocks don't block each other.
Now in your code the notStatic method doesn't synchronize on anything where as the callStatic synchronizes on the static i Integer object. So they don't block each other.
I was reading through Java Memory model and was playing with volatile. I wanted to check how Volatile will work in tandem with ThreadLocal. As per definition ThreadLocal has its own, independently initialized copy of the variable whereas when you use volatile keyword then JVM guarantees that all writes and subsequent reads are done directly from the memory. Based on the high level definitions i knew what i was trying to do will give unpredictable results. But just out of curiosity wanted to ask if someone can explain in more details as if what is going on in the background. Here is my code for your reference...
public class MyMainClass {
public static void main(String[] args) throws InterruptedException {
ThreadLocal<MyClass> local = new ThreadLocal<>();
local.set(new MyClass());
for(int i=0;i<5; i++){
Thread thread = new Thread(local.get());
thread.start();
}
}
}
public class MyClass implements Runnable {
private volatile boolean flag = false;
public void printNameTillFlagIsSet(){
if(!flag)
System.out.println("Flag is on for : " + Thread.currentThread().getName());
else
System.out.println("Flag is off for : " + Thread.currentThread().getName());
}
#Override
public void run() {
printNameTillFlagIsSet();
this.flag = true;
}
}
In your code you create a ThreadLocal reference as a local variable of your main method. You then store an instance of MyClass in it and then give that same reference of MyClass to 5 threads created in the main method.
The resulting output of the program is unpredictable since the threads are not synchronized against each other. At least one thread will see the flag as false the other four could see the flag as either true or false depending on how the thread execution is scheduled by the OS. It is possible that all 5 threads could see the flag as false, or 1 could see it false and 4 see it true or anything in between.
The use of a ThreadLocal has no impact on this run at all based on the way you are using it.
As most have pointed out you have deeply misunderstood ThreadLocal. This is how I would write it to be more accurate.
public class MyMainClass {
private static final ThreadLocal<MyClass> local = new ThreadLocal<>(){
public MyClass initialValue(){
return new MyClass();
}
}
public static void main(String[] args) throws InterruptedException {
local.set(new MyClass());
for(int i=0;i<5; i++){
Thread thread = new Thread(new Runnable(){
public void run(){
local.get().printNameTillFlagIsSet();
local.get().run();
local.get().printNameTillFlagIsSet();
}
});
thread.start();
}
}
}
So here five different instances of MyClass are created. Each thread will have their own accessible copy of each MyClass. That is Thread created at i = 0 will always have a different instance of MyClass then i = 1,2,3,4 despite how many local.get() are done.
The inner workings are a bit complicated but it can be done similar to
ConcurrentMap<Long,Thread> threadLocalMap =...;
public MyClass get(){
long id = Thread.currentThread().getId();
MyClass value = threadLocalMap.get(id);
if(value == null){
value = initialValue();
threadLocalMap.put(id,value);
}
return value;
}
To further answer your question about the volatile field. It is in essence useless here. Since the field itself is 'thread-local' there will be no ordering/memory issues that can occur.
Just don't divinize the JVM. ThreadLocal is a regular class. Inside it uses a map from current thread ID into an object instance. So that the same ThreadLocal variable could have its own value for each thread. That's all. Your variable exists only in the main thread, so it doesn't make any sence.
The volatile is something about java code optimization, It just stops all possible optimizations which allow avoid redundant memory reads/writes and execution sequence re-orderings. It is important for expecting some particular behaviour in multi-threaded environment.
You have two big problems:
1) As many pointed out, you are not using ThreadLocal properly so you don't actually have any "thread local" variables.
2) Your code is equivalent to:
MyClass someInstance = new Class();
for (...)
... new Thread(someInstance);
so you should expect to see 1 on and 4 off. However your code is badly synchronized, so you get random results. The problem is that although you declare flag as volatile, this is not enough for good synchronization since you do the check on flag in printNameTillFlagSet and then change the flag value just after that method call in run. There is a gap here where many threads can see the flag as true. You should check the flag value and change it within a synchronized block.
You main thread where you have a ThreadLocal object is being passed to all the threads. So the same instance is being passed.
So its as good as
new Thread(new MyClass());
What you could try is have an object being called by different threads with a thread local variable. This will be a proper test for ThreadLocal where each thread will get its own instance of the variable.