JAVA Threads
What will be the output?class A extends Thread{ public void run(){ for(int i=0; i<2; i++){ System.out.println(i); } }}public class Test{ public static void main(String argv[]){ Test t = new Test(); t.check(new A(){}); } public void check(A a){ a.start(); }}

None of these
0 0
Compilation succeed but runtime exception
Compilation error, class A has no start method
0 1

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JAVA Threads
Analyze the following code:public class Test implements Runnable{ public static void main(String[] args){ Test t = new Test(); t.start(); } public void run() { }}

The program does not compile because the start() method is not defined in the Test class.
The program compiles and runs fine.
The program compiles, but it does not run because the run() method is not implemented.
The program compiles, but it does not run because the start() method is not defined.

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JAVA Threads
What will happen when you attempt to compile and run the following code?1. public class Test extends Thread{2. public static void main(String argv[]){3. Test t = new Test();4. t.run();5. t.start();6. }7. public void run(){8. System.out.println("run-test");9. }10. }

Compilation fails due to an error on line 7
run-test run-test
Compilation fails due to an error on line 4
None of these
run-test

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JAVA Threads
Predict the output:class A implements Runnable{ public void run(){ try{ for(int i=0;i<4;i++){ Thread.sleep(100); System.out.println(Thread.currentThread().getName()); } }catch(InterruptedException e){ } }}public class Test{ public static void main(String argv[]) throws Exception{ A a = new A(); Thread t = new Thread(a, "A"); Thread t1 = new Thread(a, "B"); t.start(); t.join(); t1.start(); }}

None of these
A B A B A B A B
A A A A B B B B
Compilation succeed but Runtime Exception
Output order is not guaranteed

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JAVA Threads
Analyze the following code:public abstract class Test implements Runnable{ public void doSomething() { };}

The program will not compile because it does not implement the run() method.
None of these
The program compiles fine.
The program will not compile because it does not contain abstract methods.

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