What will be the output of the program?

int i = 1, j = 10; 
do 
{
    if(i++ > --j) /* Line 4 */
    {
        continue; 
    } 
} while (i < 5); 
System.out.println("i = " + i + "and j = " + j); /* Line 9 */

(1), (3), and (6) are correct. char c1 = 064770; is an octal representation of the integer value 27128, which is legal because it fits into an unsigned 16-bit integer. char c3 = 0xbeef; is a hexadecimal representation of the integer value 48879, which fits into an unsigned 16-bit integer. char c6 = '\uface'; is a Unicode representation of a character.

char c2 = 'face'; is wrong because you can't put more than one character in a char literal. The only other acceptable char literal that can go between single quotes is a Unicode value, and Unicode literals must always start with a '\u'.

char c4 = \u0022; is wrong because the single quotes are missing.

char c5 = '\iface'; is wrong because it appears to be a Unicode representation (notice the backslash), but starts with '\i' rather than '\u'.

The code doesn't compile because the method GetResult() in class A is final and so cannot be overridden.

When z == 0 , case x-2 is matched. When z == 1, case x-1 is matched and then the break occurs. When z == 2, case x, then default, then x-1 are all matched. When z == 3, default, then x-1 are matched. The rules for default are that it will fall through from above like any other case (for instance when z == 2), and that it will match when no other cases match (for instance when z==3).

The Vector.elements method returns an Enumeration over the elements of the vector. Vector implements the List interface and extends AbstractList so it is also possible to get an Iterator over a Vector by invoking the iterator or listIterator method.

A is wrong. A try statement can exist without catch, but it must have a finally statement.

B is wrong. A try statement executes a block. If a value is thrown and the try statement has one or more catch clauses that can catch it, then control will be transferred to the first such catch clause. If that catch block completes normally, then the try statement completes normally.

C is wrong. Exceptions of type Error and RuntimeException do not have to be caught, only checked exceptions (java.lang.Exception) have to be caught. However, speaking of Exceptions, Exceptions do not have to be handled in the same method as the throw statement. They can be passed to another method.

If you put a finally block after a try and its associated catch blocks, then once execution enters the try block, the code in that finally block will definitely be executed except in the following circumstances:

1. An exception arising in the finally block itself.
2. The death of the thread.
3. The use of System.exit()
4. Turning off the power to the CPU.

I suppose the last three could be classified as VM shutdown.

Option A is correct. Because assertions may be disabled, programs must not assume that the boolean expressions contained in assertions will be evaluated. Thus these expressions should be free of side effects. That is, evaluating such an expression should not affect any state that is visible after the evaluation is complete. Although it is not illegal for a boolean expression contained in an assertion to have a side effect, it is generally inappropriate, as it could cause program behaviour to vary depending on whether assertions are enabled or disabled.

Assertion checking may be disabled for increased performance. Typically, assertion checking is enabled during program development and testing and disabled for deployment.

Option B is wrong. Because you assert that something is "true". True is Boolean. So, an expression must evaluate to Boolean, not int or byte or anything else. Use the same rules for an assertion expression that you would use for a while condition.

Option C is wrong. Usually, enforcing a precondition on a public method is done by condition-checking code that you write yourself, to give you specific exceptions.

Option D is wrong. "You're never supposed to handle an assertion failure"

Not all legal uses of assertions are considered appropriate. As with so much of Java, you can abuse the intended use for assertions, despite the best efforts of Sun's Java engineers to discourage you. For example, you're never supposed to handle an assertion failure. That means don't catch it with a catch clause and attempt to recover. Legally, however, AssertionError is a subclass of Throwable, so it can be caught. But just don't do it! If you're going to try to recover from something, it should be an exception. To discourage you from trying to substitute an assertion for an exception, the AssertionError doesn't provide access to the object that generated it. All you get is the String message.

String objects are immutable. The object s above is set to "ABC". Now ask yourself if this object is changed and if so where - remember strings are immutable.

Line 2 returns a string object but does not change the originag string object s, so after line 2 s is still "ABC".

So what's happening on line 3? Java will treat line 3 like the following:

s = new StringBuffer().append(s).append("def").toString();

This effectively creates a new String object and stores its reference in the variable s, the old String object containing "ABC" is no longer referenced by a live thread and becomes available for garbage collection.

Option D is correct, compilation fails - The return type of getLength( ) in the super class is an object of reference type Integer and the return type in the sub class is an object of reference type Long. In other words, it is not an override because of the change in the return type and it is also not an overload because the argument list has not changed.

Option C is correct. HashSet implements the Set interface and the Set interface specifies collection that contains no duplicate elements.

Option A is wrong. HashSet makes no guarantees as to the iteration order of the set; in particular, it does not guarantee that the order will remain constant over time.

Option B is wrong. The set can be modified.

Option D is wrong. This is a Set and not a Map.

In the above code the array reference variable x has been declared but it has not been instantiated i.e. the new statement is missing, for example:

private static int[]x = new int[5];

private static int[x] declares a static i.e. class level array.

the "new" keyword is the word that actually creates said array.

int[5] in association with the new sets the size of the array. so since the above code contains no new or size decalarations when you try and access x[0] you are trying to access a member of an array that has been declared but not intialized hence you get a NullPointerException at runtime.