What will be the output of the program?

class Q207 
{ 
    public static void main(String[] args) 
    {
        int i1 = 5; 
        int i2 = 6; 
        String s1 = "7"; 
        System.out.println(i1 + i2 + s1); /* Line 8 */
    } 
}

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.

Math.round() adds .5 to the argument then performs a floor(). Since the code adds an additional .5 before round() is called, it's as if we are adding 1 then doing a floor(). The values that start out as integer values will in effect be incremented by 1 on the round() side but not on the ceil() side, and the noninteger values will end up equal.

You have two different threads that share one reference to a common object.

The updating and output takes place inside synchronized code.

One thread will run to completion printing the numbers 1-10.

The second thread will then run to completion printing the numbers 11-20.

Option B is correct - The notify method only wakes the thread. It does not guarantee that the thread will run.

Option A is incorrect - just because another thread activates the modify method in A this does not mean that the thread will automatically resume execution

Option C is incorrect - This is incorrect because as said in Answer B notify only wakes the thread but further to this once it is awake it goes back into the stack and awaits execution therefore it is not a "direct and sole consequence of the notify call"

Option D is incorrect - The notify method wakes one waiting thread up. If there are more than one sleeping threads then the choice as to which thread to wake is made by the machine rather than you therefore you cannot guarantee that the notify'ed thread will be the first waiting thread.