多线程执行顺序

Question

Let's say we have this scenario:

class Stack{

public void main{

ChildThread1 t1 = new ChildThread1;
ChildThread1 t2 = new ChildThread1;
ChildThread1 t3 = new ChildThread1;

//then we make some ChildThread2 objects and some ChildThread3 objects

ChildThread2 s1 = new ChildThread2;
//...

ChildThread3 v1 = new ChildThread3;
//...

//now we let all threads start in mix order
t1.start();
v1.start();
//...

SOP("All threads are ready");

//then we let them run with join()
t1.join();
t2.join();
t3.join();

s1.join();
//...

v1.join();
//...

Each type of thread prints its own unique statement when it is running.

I noticed that every time I execute the program, the output is always different. For example, a statement from ChilThread1 t1 will print in the middle of output instead of beginning (since t1 starts first) or the statement "all threads are ready" will pop in middle of thread execution (Example: ChildThread2 is 'all threads are ready' running )

So I tried to find an answer and I found this site: http://www.avajava.com/tutorials/lessons/how-do-i-use-threads-join-method.html The site basically says there's no guarantee order of execution when you use start()

So am I right to assume that this weird order of prints is because start() does not guarantee order of execution? Will this reason also apply to 'all threads are ready' problem?

Answer 1

The whole point of threads is that they can be executed concurrently. If you want to ensure specific order in which things are done, you either have to move away from using threads, or use explicit synchronization.

So am I right to assume that this weird order of prints is because start() does not guarantee order of execution?

That's right. When you start a thread, there's basically a race condition between the main thread and the newly created thread. This means that nothing can be said about the relative order in which things happen between the two threads. If you want to ensure specific ordering, use synchronization.

Answer 2

Easy way to maintain ordering in Thread Execution is to use Semaphore

public class Semaphore {

int value;

public Semaphore(int intialValue) {
    this.value = intialValue;
}

public synchronized void p() {
    while (value <= 0) {
        try {
            this.wait();
        } catch (InterruptedException e) {
        }

    }
    value = value - 1;
}

public synchronized void v() {
    value = value + 1;
    this.notify();
}

}

public class ThreadSync {

static Semaphore semaphore = new Semaphore(0);

public static void main(String[] args) {
    // t1 should be executed before t2
    Thread t1 = new Thread(new Runnable() {

        @Override
        public void run() {
            semaphore.p();
            System.out.println("executing " + Thread.currentThread().getName());
        }
    });
    Thread t2 = new Thread(new Runnable() {

        @Override
        public void run() {
            System.out.println("executing " + Thread.currentThread().getName());
            semaphore.v();
        }
    });

    t1.setName("Thread 1");
    t2.setName("Thread 2");
    t2.start();
    t1.start();

}

}

Answer 3

So am I right to assume that this weird order of prints is because start() does not guarantee order of execution?

Yes. You are right.

Will this reason also apply to 'all threads are ready' problem?

Yes. Right again. Your SOP is run by main thread. So, it's possible that t1 can print something before main gets chance to execute it's SOP .

Answer 4

When you start a thread, the started thread runs in parallel of all the already running threads. The thread scheduler dispatches the various threads on the available processors, and each thread gets some processor time, each in his turn. But the processor, the order and the time assigned to each thread is up to the OS thread scheduler, and you have absolutely no guarantee.