问题
In the following code, if i use sysout statement inside for loop then the code executes and goes inside the loop after the condition met but if i do not use sysout statement inside loop then then infinite loop goes on without entering inside the if condition even if the if condition is satisfied.. can anyone please help me to find out the exact reason for this. Just A sysout statement make the if condition to become true. why is it so?
The code is as follows:-
class RunnableDemo implements Runnable {
private Thread t;
private String threadName;
RunnableDemo( String name){
threadName = name;
System.out.println("Creating " + threadName );
}
public void run() {
System.out.println("Running " + threadName );
for(;;)
{
//Output 1: without this sysout statement.
//Output 2: After uncommenting this sysout statement
//System.out.println(Thread.currentThread().isInterrupted());
if(TestThread.i>3)
{
try {
for(int j = 4; j > 0; j--) {
System.out.println("Thread: " + threadName + ", " + j);
}
} catch (Exception e) {
System.out.println("Thread " + threadName + " interrupted.");
}
System.out.println("Thread " + threadName + " exiting.");
}
}
}
public void start ()
{
System.out.println("Starting " + threadName );
if (t == null)
{
t = new Thread (this, threadName);
t.start ();
}
}
}
public class TestThread {
static int i=0;
public static void main(String args[]) {
RunnableDemo R1 = new RunnableDemo( "Thread-1");
R1.start();
try {
Thread.sleep(10000);
} catch (InterruptedException e) {
e.printStackTrace();
}
i+=4;
System.out.println(i);
}
}
Output without sysout statement in the infinite loop:- enter image description here
Output with sysout statement in the infinite loop:- enter image description here
回答1:
The problem here can be fixed by changing
static int i=0;
to
static volatile int i=0;
Making a variable volatile has a number of complex consequences and I am not an expert at this. So, I'll try to explain how I think about it.
The variable i lives in your main memory, your RAM. But RAM is slow, so your processor copies it to the faster (and smaller) memory: the cache. Multiple caches in fact, but thats irrelevant.
But when two threads on two different processors put their values in different caches, what happens when the value changes? Well, if thread 1 changes the value in cache 1, thread 2 still uses the old value from cache 2. Unless we tell both threads that this variable i might be changing at any time as if it were magic. That's what the volatile keyword does.
So why does it work with the print statement? Well, the print statement invokes a lot of code behind the scenes. Some of this code most likely contains a synchronized block or another volatile variable, which (by accident) also refreshes the value of i in both caches. (Thanks to Marco13 for pointing this out).
Next time you try to access i, you get the updated value!
PS: I'm saying RAM here, but its probably the closest shared memory between the two threads, which could be a cache if they are hyperthreaded for instance.
This is a great explanation too (with pictures!):
http://tutorials.jenkov.com/java-concurrency/volatile.html
回答2:
When you are accessing a variable value, the changes aren't written to (or loaded from) the actual memory location every time. The value can be loaded into a CPU register, or cached, and sit there until the caches are flushed. Moreover, because TestThread.i is not modified inside the loop at all, the optimizer might decide to just replace it with a check before the loop, and get rid of the if statement entirely (I do not think it is actually happening in your case, but the point is that it might).
The instruction that makes the thread to flush its caches and synchronize them with the current contents of physical memory is called memory barrier. There are two ways in Java to force a memory barrier: enter or exit a synchronized block or access a volatile variable.
When either of those events happens, the cached are flushed, and the thread is guaranteed to both see an up-to-date view of the memory contents, and have all the changes it has made locally committed to memory.
So, as suggested in comments, if your declare TestThread.i as volatile, the problem will go away, because whenever the value is modified, the change will be committed immediately, and the optimizer will know not to optimizer,e the check away from the loop, and not to cache the value.
Now, why does adding a print statement changes the behaviour? Well, there is a lot of synchronization going on inside the io, the thread hits a memory barrier somewhere, and loads the fresh value. This is just a coincidence.
来源：https://stackoverflow.com/questions/34655947/delay-in-running-thread-due-to-system-out-println-statement