What will be the output of the program?

public class WrapTest 
{
    public static void main(String [] args) 
    {
        int result = 0;
        short s = 42;
        Long x = new Long("42");
        Long y = new Long(42);
        Short z = new Short("42");
        Short x2 = new Short(s);
        Integer y2 = new Integer("42");
        Integer z2 = new Integer(42);

        if (x == y) /* Line 13 */
            result = 1;
        if (x.equals(y) ) /* Line 15 */
            result = result + 10;
        if (x.equals(z) ) /* Line 17 */
            result = result + 100;
        if (x.equals(x2) ) /* Line 19 */
            result = result + 1000;
        if (x.equals(z2) ) /* Line 21 */
            result = result + 10000;

        System.out.println("result = " + result);
    }
}

Correct Answer: The notifyAll() method must be called from a synchronized context.

Explanation:

class Test1 
{
    public int value;
    public int hashCode() { return 42; }
}
class Test2 
{
    public int value;
    public int hashcode() { return (int)(value^5); }
}

Correct Answer: The Test1 hashCode() method is less efficient than the Test2 hashCode() method.

Explanation:

x = 0;
if (x1.hashCode() != x2.hashCode() )  x = x + 1;
if (x3.equals(x4) )  x = x + 10;
if (!x5.equals(x6) ) x = x + 100;
if (x7.hashCode() == x8.hashCode() )  x = x + 1000;
System.out.println("x = " + x);

Correct Answer: x3.hashCode() == x4.hashCode()

Explanation:

By contract, if two objects are equivalent according to the equals() method, then the hashCode() method must evaluate them to be ==.

Option A is incorrect because if the hashCode() values are not equal, the two objects must not be equal.

Option C is incorrect because if equals() is not true there is no guarantee of any result from hashCode().

Option D is incorrect because hashCode() will often return == even if the two objects do not evaluate to equals() being true.

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);
        }
    }
}

Correct Answer: This code will not compile.

Explanation:

D. The sleep() method must be enclosed in a try/catch block, or the method printAll() must declare it throws the InterruptedException.

A is incorrect, but it would be correct if the InterruptedException was dealt with.

B is incorrect, but it would still be incorrect if the InterruptedException was dealt with because all Java code, including the main() method, runs in threads.

C is incorrect. The sleep() method is static, so even if it is called on an instance, it still always affects the currently executing thread.

public class MyProgram 
{
    public static void throwit() 
    {
        throw new RuntimeException();
    }
    public static void main(String args[])
    {
        try 
        {
            System.out.println("Hello world ");
            throwit();
            System.out.println("Done with try block ");
        }
        finally 
        {
            System.out.println("Finally executing ");
        }
    }
}

Correct Answer: The program will print Hello world, then will print Finally executing, then will print that a RuntimeException has occurred.

Explanation:

Once the program throws a RuntimeException (in the throwit() method) that is not caught, the finally block will be executed and the program will be terminated. If a method does not handle an exception, the finally block is executed before the exception is propagated.

interface Base 
{
    boolean m1 ();
    byte m2(short s);
}

Correct Answer: 3 and 4

Explanation:

(3) is correct because an abstract class doesn't have to implement any or all of its interface's methods. (4) is correct because the method is correctly implemented ((7 > 4) is a boolean).

(1) is incorrect because interfaces don't implement anything. (2) is incorrect because classes don't extend interfaces. (5) is incorrect because interface methods are implicitly public, so the methods being implemented must be public.

public class MyOuter 
{
    public static class MyInner 
    {
        public static void foo() { }
    }
}

Correct Answer: MyOuter.MyInner m = new MyOuter.MyInner();

Explanation:

MyInner is a static nested class, so it must be instantiated using the fully-scoped name of MyOuter.MyInner.

Option B is incorrect because it doesn't use the enclosing name in the new.

Option C is incorrect because it uses incorrect syntax. When you instantiate a nested class by invoking new on an instance of the enclosing class, you do not use the enclosing name. The difference between Option A and C is that Option C is calling new on an instance of the enclosing class rather than just new by itself.

Option D is incorrect because it doesn't use the enclosing class name in the variable declaration.

public class F0091 
{    
    public void main( String[] args ) 
    {  
        System.out.println( "Hello" + args[0] ); 
    } 
}

Correct Answer: The code does not run.

Explanation:

Option D is correct. A runtime error will occur owning to the main method of the code fragment not being declared static:

Exception in thread "main" java.lang.NoSuchMethodError: main

The Java Language Specification clearly states: "The main method must be declared public, static, and void. It must accept a single argument that is an array of strings."

class Bar { } 
class Test 
{  
    Bar doBar() 
    {
        Bar b = new Bar(); /* Line 6 */
        return b; /* Line 7 */
    } 
    public static void main (String args[]) 
    { 
        Test t = new Test();  /* Line 11 */
        Bar newBar = t.doBar();  /* Line 12 */
        System.out.println("newBar"); 
        newBar = new Bar(); /* Line 14 */
        System.out.println("finishing"); /* Line 15 */
    } 
}

Correct Answer: after line 14

Explanation:

Option B is correct. All references to the Bar object created on line 6 are destroyed when a new reference to a new Bar object is assigned to the variable newBar on line 14. Therefore the Bar object, created on line 6, is eligible for garbage collection after line 14.

Option A is wrong. This actually protects the object from garbage collection.

Option C is wrong. Because the reference in the doBar() method is returned on line 7 and is stored in newBar on line 12. This preserver the object created on line 6.

Option D is wrong. Not applicable because the object is eligible for garbage collection after line 14.

public class ExceptionTest 
{ 
    class TestException extends Exception {} 
    public void runTest() throws TestException {} 
    public void test() /* Point X */ 
    { 
        runTest(); 
    } 
}

Correct Answer: throws Exception

Explanation:

Option B is correct. This works because it DOES throw an exception if an error occurs.

Option A is wrong. If you compile the code as given the compiler will complain:

"unreported exception must be caught or declared to be thrown" The class extends Exception so we are forced to test for exceptions.

Option C is wrong. The catch statement belongs in a method body not a method specification.

Option D is wrong. TestException is a subclass of Exception therefore the test method, in this example, must throw TestException or some other class further up the Exception tree. Throwing RuntimeException is just not on as this belongs in the java.lang.RuntimeException branch (it is not a superclass of TestException). The compiler complains with the same error as in A above.