Exercise: Threads

Questions for: Threads

The following block of code creates a Thread using a Runnable target: 
Runnable target = new MyRunnable();
Thread myThread = new Thread(target);
Which of the following classes can be used to create the target, so that the preceding code compiles correctly?
A:
public class MyRunnable extends Runnable{public void run(){}}
B:
public class MyRunnable extends Object{public void run(){}}
C:
public class MyRunnable implements Runnable{public void run(){}}
D:
public class MyRunnable implements Runnable{void run(){}}
Answer: C

The class correctly implements the Runnable interface with a legal public void run() method.

Option A is incorrect because interfaces are not extended; they are implemented.

Option B is incorrect because even though the class would compile and it has a valid public void run() method, it does not implement the Runnable interface, so the compiler would complain when creating a Thread with an instance of it.

Option D is incorrect because the run() method must be public.

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Which two statements are true?
  1. Deadlock will not occur if wait()/notify() is used
  2. A thread will resume execution as soon as its sleep duration expires.
  3. Synchronization can prevent two objects from being accessed by the same thread.
  4. The wait() method is overloaded to accept a duration.
  5. The notify() method is overloaded to accept a duration.
  6. Both wait() and notify() must be called from a synchronized context.
A:
1 and 2
B:
3 and 5
C:
4 and 6
D:
1 and 3
Answer: C

Statements (4) and (6) are correct. (4) is correct because the wait() method is overloaded to accept a wait duration in milliseconds. If the thread has not been notified by the time the wait duration has elapsed, then the thread will move back to runnable even without having been notified.

(6) is correct because wait()/notify()/notifyAll() must all be called from within a synchronized, context. A thread must own the lock on the object its invoking wait()/notify()/notifyAll() on.

(1) is incorrect because wait()/notify() will not prevent deadlock.

(2) is incorrect because a sleeping thread will return to runnable when it wakes up, but it might not necessarily resume execution right away. To resume executing, the newly awakened thread must still be moved from runnable to running by the scheduler.

(3) is incorrect because synchronization prevents two or more threads from accessing the same object.

(5) is incorrect because notify() is not overloaded to accept a duration.

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Which two can be used to create a new Thread?
  1. Extend java.lang.Thread and override the run() method.
  2. Extend java.lang.Runnable and override the start() method.
  3. Implement java.lang.Thread and implement the run() method.
  4. Implement java.lang.Runnable and implement the run() method.
  5. Implement java.lang.Thread and implement the start() method.
A:
1 and 2
B:
2 and 3
C:
1 and 4
D:
3 and 4
Answer: C

There are two ways of creating a thread; extend (sub-class) the Thread class and implement the Runnable interface. For both of these ways you must implement (override and not overload) the public void run() method.

(1) is correct - Extending the Thread class and overriding its run method is a valid procedure.

(4) is correct - You must implement interfaces, and runnable is an interface and you must also include the run method.

(2) is wrong - Runnable is an interface which implements not Extends. Gives the error: (No interface expected here)

(3) is wrong - You cannot implement java.lang.Thread (This is a Class). (Implements Thread, gives the error: Interface expected). Implements expects an interface.

(5) is wrong - You cannot implement java.lang.Thread (This is a class). You Extend classes, and Implement interfaces. (Implements Thread, gives the error: Interface expected)

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What will be the output of the program? 
class MyThread extends Thread 
{
    public static void main(String [] args) 
    {
        MyThread t = new MyThread();
        Thread x = new Thread(t);
        x.start(); /* Line 7 */
    }
    public void run() 
    {
        for(int i = 0; i < 3; ++i) 
        {
            System.out.print(i + "..");
        }
    }
}
A:
Compilation fails.
B:
1..2..3..
C:
0..1..2..3..
D:
0..1..2..
Answer: D

The thread MyThread will start and loop three times (from 0 to 2).

Option A is incorrect because the Thread class implements the Runnable interface; therefore, in line 7, Thread can take an object of type Thread as an argument in the constructor.

Option B and C are incorrect because the variable i in the for loop starts with a value of 0 and ends with a value of 2.

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What will be the output of the program? 
public class Test 
{
    public static void main (String [] args) 
    {
        final Foo f = new Foo();
        Thread t = new Thread(new Runnable() 
        {
            public void run() 
            {
                f.doStuff();
            }
        });
        Thread g = new Thread() 
        {
            public void run() 
            {
                f.doStuff();
            }
        };
        t.start();
        g.start();
    }
}
class Foo 
{
    int x = 5;
    public void doStuff() 
    {
        if (x < 10) 
        {
            // nothing to do
            try 
            {
                wait();
                } catch(InterruptedException ex) { }
        } 
        else 
        {
            System.out.println("x is " + x++);
            if (x >= 10) 
            {
                notify();
            }
        }
    }
}
A:
The code will not compile because of an error on notify(); of class Foo.
B:
The code will not compile because of some other error in class Test.
C:
An exception occurs at runtime.
D:
It prints "x is 5 x is 6".
Answer: C

C is correct because the thread does not own the lock of the object it invokes wait() on. If the method were synchronized, the code would run without exception.

A, B are incorrect because the code compiles without errors.

D is incorrect because the exception is thrown before there is any output.

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