Threads - Finding the output Questions and Answers

Q: Sample66 is run - what will be the output? public class Sample66 { public static void main(String[] args) throws InterruptedException { long start = System.currentTimeMillis(); Thread.sleep(0); System.out.println(System.currentTimeMillis() - start < 100); } }

Correct answer: trueThread.sleep(0) is legal and returns essentially at once - it is at most a hint to the scheduler that the thread is willing to yield. The elapsed time is therefore a millisecond or two and the comparison prints true. Only a *negative* argument throws IllegalArgumentException.

61.

Sample61 is run - what will be the output?

public class Sample61
{
    public static void main(String[] args)
    {
        final Object lock = new Object();
        System.out.print(Thread.holdsLock(lock) + " ");
        synchronized (lock)
        {
            System.out.print(Thread.holdsLock(lock));
        }
    }
}

A.false true
B.true true
C.false false
D.true false

Answer: A

holdsLock reports whether the current thread owns the monitor of the given object at that moment. Outside the synchronized block it does not, and inside it does. The method exists for assertions and diagnostics rather than for control flow.

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62.

Sample62 is run - what will be the output?

public class Sample62
{
    static int shared = 0;

    static class Job implements Runnable
    {
        int local = 0;

        public void run()
        {
            for (int i = 0; i < 3; i++) { local++; shared++; }
        }
    }

    public static void main(String[] args) throws InterruptedException
    {
        Job job = new Job();
        Thread a = new Thread(job), b = new Thread(job);
        a.start(); a.join(); b.start(); b.join();
        System.out.println(job.local + " " + shared);
    }
}

A.3 3
B.6 6
C.3 6
D.6 3

Answer: B

Both threads run the same Job instance, so its field is shared just as the static one is; nothing is copied per thread. Each thread adds three to both, giving six each. Starting the second thread only after joining the first removes any interleaving, so the totals are exact.

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63.

Sample63 is run - what will be the output?

public class Sample63
{
    static class Broken
    {
        Broken()
        {
            new Thread(() -> System.out.println(value)).start();
            value = 42;
        }

        int value;
    }

    public static void main(String[] args) throws InterruptedException
    {
        new Broken();
        Thread.sleep(200);
    }
}

A.0 always, because the field is read before it is assigned
B.A compile error, because a constructor may not start a thread
C.Either 0 or 42, because the reference escaped before construction finished
D.42 always, because the constructor completes before the thread runs

Answer: C

Starting a thread from inside the constructor publishes this before the object is finished, so the new thread may observe the field at its default value or at 42 depending on timing. Publishing a reference only after construction has completed is what avoids the race.

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64.

Sample64 is run - what will be the output?

public class Sample64
{
    public static void main(String[] args) throws InterruptedException
    {
        Thread low = new Thread(() -> System.out.print("L"));
        Thread high = new Thread(() -> System.out.print("H"));
        low.setPriority(Thread.MIN_PRIORITY);
        high.setPriority(Thread.MAX_PRIORITY);
        low.start(); high.start();
        low.join(); high.join();
    }
}

A.HL always
B.LH always
C.An IllegalArgumentException from setPriority
D.LH or HL; priority does not decide the order

Answer: D

Priority is a hint to the platform's scheduler and its effect varies between JVMs and operating systems, so neither ordering is promised. 1 and 10 are legal values, so nothing is thrown. Real ordering needs join between the two starts, a lock or a latch.

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65.

Sample65 is run - what will be the output?

public class Sample65
{
    static final Runnable POLITE = () ->
    {
        for (int spin = 0; spin < 3; spin++)
        {
            Thread.yield();
        }
        System.out.print("worker");
    };

    public static void main(String[] args) throws InterruptedException
    {
        Thread courteous = new Thread(POLITE);
        courteous.start();
        courteous.join();
        System.out.print(" main");
    }
}

A.worker main
B.main worker
C.Either order
D.worker only

Answer: A

yield is only a hint that the scheduler may ignore, and it never ends a thread or transfers control to a named one, however many times it is called. Whatever it does or does not do, the join still guarantees the worker finishes before main's print, so the order is fixed by the join alone.

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66.

Sample66 is run - what will be the output?

public class Sample66
{
    public static void main(String[] args) throws InterruptedException
    {
        long start = System.currentTimeMillis();
        Thread.sleep(0);
        System.out.println(System.currentTimeMillis() - start < 100);
    }
}

A.The program blocks for ever.
B.true
C.false
D.An IllegalArgumentException

Answer: B

Correct answer: true

Thread.sleep(0) is legal and returns essentially at once - it is at most a hint to the scheduler that the thread is willing to yield. The elapsed time is therefore a millisecond or two and the comparison prints true. Only a *negative* argument throws IllegalArgumentException.

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67.

Sample67 is run - what will be the output?

import java.util.concurrent.Semaphore;

public class Sample67
{
    public static void main(String[] args) throws InterruptedException
    {
        Semaphore permits = new Semaphore(1);
        permits.acquire();
        System.out.print(permits.availablePermits() + " ");
        permits.release();
        System.out.print(permits.availablePermits());
    }
}

A.1 1
B.0 0
C.0 1
D.1 0

Answer: C

acquire takes a permit and blocks if none is available, so the count drops to zero; release returns one, taking it back to one. A semaphore counts rather than tracking an owner, which is why release may be called by a thread that never acquired.

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68.

Sample68 is run - what will be the output?

import java.util.*;
import java.util.concurrent.*;

public class Sample68
{
    public static void main(String[] args) throws Exception
    {
        ExecutorService pool = Executors.newFixedThreadPool(3);
        List<Callable<String>> jobs = Arrays.asList(() -> "a", () -> "b");
        List<Future<String>> done = pool.invokeAll(jobs);
        for (Future<String> f : done) System.out.print(f.get());
        pool.shutdown();
    }
}

A.ba
B.Either ab or ba
C.aabb
D.ab

Answer: D

invokeAll returns only when every task has completed, and the list of futures comes back in the same order as the tasks were supplied, whatever order they actually ran in. So iterating the list gives a and then b deterministically.

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69.

Sample69 is run - what will be the output?

import java.util.concurrent.atomic.AtomicInteger;

public class Sample69
{
    public static void main(String[] args)
    {
        AtomicInteger n = new AtomicInteger(5);
        System.out.print(n.compareAndSet(5, 9) + " ");
        System.out.print(n.compareAndSet(5, 12) + " ");
        System.out.print(n.get());
    }
}

A.true false 9
B.true true 12
C.false false 5
D.true false 12

Answer: A

The first call finds the expected 5 and writes 9, reporting success. The second still expects 5, but the value is now 9, so it writes nothing and returns false. The final value is therefore 9, and the boolean result is what lets a retry loop work without a lock.

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70.

Sample70 is run - what will be the output?

import java.util.*;

public class Sample70
{
    public static void main(String[] args) throws InterruptedException
    {
        List<Integer> list = Collections.synchronizedList(new ArrayList<>());
        Runnable job = () -> { for (int i = 0; i < 500; i++) list.add(i); };
        Thread a = new Thread(job), b = new Thread(job);
        a.start(); b.start(); a.join(); b.join();
        System.out.println(list.size());
    }
}

A.A ConcurrentModificationException from one of the threads
B.1000
C.500
D.An unpredictable value below 1000

Answer: B

Each add is individually synchronized, and a list keeps duplicates, so all 1000 additions land and the size is exact. What the wrapper does not make atomic is a sequence of calls, so iterating or a check-then-add would still need the caller to hold the lock.

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