What will be the output of the program? class MyThread extends Thread { public static void main(String [] args) { MyThread t = new MyThread(); t.start(); System.out.print("one. "); t.start(); System.out.print("two. "); } public void run() { System.out.print("Thread "); } }

Compilation fails
An exception occurs at runtime.
It prints "Thread one. Thread two."
The output cannot be determined.

Correct Answer: An exception occurs at runtime.

Explanation:

When the start() method is attempted a second time on a single Thread object, the method will throw an IllegalThreadStateException (you will not need to know this exception name for the exam). Even if the thread has finished running, it is still illegal to call start() again.

← Previous Question Next Question→

More Questions from Threads

What will be the output of the program? class s1 implements Runnable { int x = 0, y = 0; int addX() {x++; return x;} int addY() {y++; return y;} public void run() { for(int i = 0; i < 10; i++) System.out.println(addX() + " " + addY()); } public static void main(String args[]) { s1 run1 = new s1(); s1 run2 = new s1(); Thread t1 = new Thread(run1); Thread t2 = new Thread(run2); t1.start(); t2.start(); } }
What will be the output of the program? class s implements Runnable { int x, y; public void run() { for(int i = 0; i < 1000; i++) synchronized(this) { x = 12; y = 12; } System.out.print(x + " " + y + " "); } public static void main(String args[]) { s run = new s(); Thread t1 = new Thread(run); Thread t2 = new Thread(run); t1.start(); t2.start(); } }
The static method Thread.currentThread() returns a reference to the currently executing Thread object. What is the result of this code? class Test { public static void main(String [] args) { printAll(args); } public static void printAll(String[] lines) { for(int i = 0; i < lines.length; i++) { System.out.println(lines[i]); Thread.currentThread().sleep(1000); } } }
What will be the output of the program? public class ThreadTest extends Thread { public void run() { System.out.println("In run"); yield(); System.out.println("Leaving run"); } public static void main(String []argv) { (new ThreadTest()).start(); } }
What will be the output of the program? class s1 extends Thread { public void run() { for(int i = 0; i < 3; i++) { System.out.println("A"); System.out.println("B"); } } } class Test120 extends Thread { public void run() { for(int i = 0; i < 3; i++) { System.out.println("C"); System.out.println("D"); } } public static void main(String args[]) { s1 t1 = new s1(); Test120 t2 = new Test120(); t1.start(); t2.start(); } }
What will be the output of the program? class MyThread extends Thread { MyThread() { System.out.print(" MyThread"); } public void run() { System.out.print(" bar"); } public void run(String s) { System.out.println(" baz"); } } public class TestThreads { public static void main (String [] args) { Thread t = new MyThread() { public void run() { System.out.println(" foo"); } }; t.start(); } }
Which two statements are true? Deadlock will not occur if wait()/notify() is used A thread will resume execution as soon as its sleep duration expires. Synchronization can prevent two objects from being accessed by the same thread. The wait() method is overloaded to accept a duration. The notify() method is overloaded to accept a duration. Both wait() and notify() must be called from a synchronized context.
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?
Which statement is true?
Which statement is true?

Discussion & Comments

No comments yet. Be the first to comment!

Join Discussion