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Java Questions And Answers Sample Test 10


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Java Test 10


You have scored 3 out of 10
You Scored:2
Total Questions:2
Attended Questions:2
Correct Answered:2
1. Which is a reserved word in the Java programming language?



Explanation:

The word "native" is a valid keyword, used to modify a method declaration.

Option A, D and E are not keywords. Option C is wrong because the keyword for subclassing in Java is extends, not 'subclasses'.


2.
What will be the output of the program?

class PassA 
{
    public static void main(String [] args) 
    {
        PassA p = new PassA();
        p.start();
    }

    void start() 
    {
        long [] a1 = {3,4,5};
        long [] a2 = fix(a1);
        System.out.print(a1[0] + a1[1] + a1[2] + " ");
        System.out.println(a2[0] + a2[1] + a2[2]);
    }

    long [] fix(long [] a3) 
    {
        a3[1] = 7;
        return a3;
    }
}



Explanation:

Output: 15 15

The reference variables a1 and a3 refer to the same long array object. When the [1] element is updated in the fix() method, it is updating the array referred to by a1. The reference variable a2 refers to the same array object.

So Output: 3+7+5+" "3+7+5

Output: 15 15 Because Numeric values will be added


3.
public class ExceptionTest 
    class TestException extends Exception {} 
    public void runTest() throws TestException {} 
    public void test() /* Point X */ 
    { 
        runTest(); 
    } 
}
At Point X on line 5, which code is necessary to make the code compile?



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.


4. Which collection class allows you to grow or shrink its size and provides indexed access to its elements, but whose methods are not synchronized?



Explanation:

All of the collection classes allow you to grow or shrink the size of your collection. ArrayList provides an index to its elements. The newer collection classes tend not to have synchronized methods. Vector is an older implementation of ArrayList functionality and has synchronized methods; it is slower than ArrayList.


5.
What will be the output of the program?

class SSBool 
{
    public static void main(String [] args) 
    {
        boolean b1 = true;
        boolean b2 = false;
        boolean b3 = true;
        if ( b1 & b2 | b2 & b3 | b2 ) /* Line 8 */
            System.out.print("ok ");
        if ( b1 & b2 | b2 & b3 | b2 | b1 ) /*Line 10*/
            System.out.println("dokey");
    }
}



Explanation:

The & operator has a higher precedence than the | operator so that on line 8 b1 and b2 are evaluated together as are b2 & b3. The final b1 in line 10 is what causes that if test to be true. Hence it prints "dokey".


6.
public class X 
{
    public static void main(String [] args) 
    {
        X x = new X();
        X x2 = m1(x); /* Line 6 */
        X x4 = new X();
        x2 = x4; /* Line 8 */
        doComplexStuff();
    }
    static X m1(X mx) 
    {
        mx = new X();
        return mx;
    }
}
After line 8 runs. how many objects are eligible for garbage collection?



Explanation:

By the time line 8 has run, the only object without a reference is the one generated as a result of line 6. Remember that "Java is pass by value," so the reference variable x is not affected by the m1() method.


7.
What will be the output of the program?

public class Foo 
{
    Foo() 
    {
        System.out.print("foo");
    }
    
class Bar
{
    Bar() 
    {
        System.out.print("bar");
    }
    public void go() 
    {
        System.out.print("hi");
    }
} /* class Bar ends */

    public static void main (String [] args) 
    {
        Foo f = new Foo();
        f.makeBar();
    }
    void makeBar() 
    {
        (new Bar() {}).go();
    }
}/* class Foo ends */



Explanation:

Option C is correct because first the Foo instance is created, which means the Foo constructor runs and prints "foo". Next, the makeBar() method is invoked which creates a Bar, which means the Bar constructor runs and prints "bar", and finally the go() method is invoked on the new Bar instance, which means the go() method prints "hi".


8. Which statement is true?



Explanation:

Option A is correct. The assertion status can be set for a named top-level class and any nested classes contained therein. This setting takes precedence over the class loader's default assertion status, and over any applicable per-package default. If the named class is not a top-level class, the change of status will have no effect on the actual assertion status of any class.

Option B is wrong. Is there such a thing as conditional compilation in Java?

Option C is wrong. For private methods - yes. But do not use assertions to check the parameters of a public method. An assert is inappropriate in public methods because the method guarantees that it will always enforce the argument checks. A public method must check its arguments whether or not assertions are enabled. Further, the assert construct does not throw an exception of the specified type. It can throw only an AssertionError.

Option D is wrong. Because you're never supposed to handle an assertion failure. That means don't catch it with a catch clause and attempt to recover.


9.
What will be the output of the program?

class Exc0 extends Exception { } 
class Exc1 extends Exc0 { } /* Line 2 */
public class Test 
{  
    public static void main(String args[]) 
    { 
        try 
        {  
            throw new Exc1(); /* Line 9 */
        } 
        catch (Exc0 e0) /* Line 11 */
        {
            System.out.println("Ex0 caught"); 
        } 
        catch (Exception e) 
        {
            System.out.println("exception caught");  
        } 
    } 
}



Explanation:

An exception Exc1 is thrown and is caught by the catch statement on line 11. The code is executed in this block. There is no finally block of code to execute.


10.
Assume the following method is properly synchronized and called from a thread A on an object B:

wait(2000);
After calling this method, when will the thread A become a candidate to get another turn at the CPU?



Explanation:

Option A. Either of the two events (notification or wait time expiration) will make the thread become a candidate for running again.

Option B is incorrect because a waiting thread will not return to runnable when the lock is released, unless a notification occurs.

Option C is incorrect because the thread will become a candidate immediately after notification, not two seconds afterwards.

Option D is also incorrect because a thread will not come out of a waiting pool just because a lock has been released.


11.
public void test(int x) 
    int odd = 1; 
    if(odd) /* Line 4 */
    {
        System.out.println("odd"); 
    } 
    else 
    {
        System.out.println("even"); 
    } 
}
Which statement is true?



Explanation:

The compiler will complain because of incompatible types (line 4), the if expects a boolean but it gets an integer.


12.
Which three are valid declarations of a float?

float f1 = -343;
float f2 = 3.14;
float f3 = 0x12345;
float f4 = 42e7;
float f5 = 2001.0D;
float f6 = 2.81F;



Explanation:

(1) and (3) are integer literals (32 bits), and integers can be legally assigned to floats (also 32 bits). (6) is correct because (F) is appended to the literal, declaring it as a float rather than a double (the default for floating point literals).

(2), (4),and (5) are all doubles.


13.
What will be the output of the program?

public class Switch2 
{
    final static short x = 2;
    public static int y = 0;
    public static void main(String [] args) 
    {
        for (int z=0; z < 3; z++) 
        {
            switch (z) 
            {
                case x: System.out.print("0 ");
                case x-1: System.out.print("1 ");
                case x-2: System.out.print("2 ");
            }
        }
    }
}



Explanation:

The case expressions are all legal because x is marked final, which means the expressions can be evaluated at compile time. In the first iteration of the for loop case x-2 matches, so 2 is printed. In the second iteration, x-1 is matched so 1 and 2 are printed (remember, once a match is found all remaining statements are executed until a break statement is encountered). In the third iteration, x is matched. So 0 1 and 2 are printed.


14.
Which of the following are Java reserved words?

run
import
default
implement



Explanation:

(2) - This is a Java keyword

(3) - This is a Java keyword

(1) - Is incorrect because although it is a method of Thread/Runnable it is not a keyword

(4) - This is not a Java keyword the keyword is implements


15. Which one of these lists contains only Java programming language keywords?



Explanation:

All the words in option B are among the 49 Java keywords. Although goto reserved as a keyword in Java, goto is not used and has no function.

Option A is wrong because the keyword for the primitive int starts with a lowercase i.

Option C is wrong because "virtual" is a keyword in C++, but not Java.

Option D is wrong because "constant" is not a keyword. Constants in Java are marked static and final.

Option E is wrong because "include" is a keyword in C, but not in Java.


16. Which method registers a thread in a thread scheduler?



Explanation:

Option C is correct. The start() method causes this thread to begin execution; the Java Virtual Machine calls the run method of this thread.

Option A is wrong. The run() method of a thread is like the main() method to an application. Starting the thread causes the object's run method to be called in that separately executing thread.

Option B is wrong. There is no construct() method in the Thread class.

Option D is wrong. There is no register() method in the Thread class.


17.
public class While 
{
    public void loop() 
    {
        int x= 0;
        while ( 1 ) /* Line 6 */
        {
            System.out.print("x plus one is " + (x + 1)); /* Line 8 */
        }
    }
}
Which statement is true?



Explanation:

Using the integer 1 in the while statement, or any other looping or conditional construct for that matter, will result in a compiler error. This is old C Program syntax, not valid Java.

A, B and C are incorrect because line 1 is valid (Java is case sensitive so While is a valid class name). Line 8 is also valid because an equation may be placed in a String operation as shown.


18.
public class Test 
    public void foo() 
    {
        assert false; /* Line 5 */
        assert false; /* Line 6 */
    } 
    public void bar()
    {
        while(true)
        {
            assert false; /* Line 12 */
        } 
        assert false;  /* Line 14 */
    } 
}
What causes compilation to fail?



Explanation:

Option D is correct. Compilation fails because of an unreachable statement at line 14. It is a compile-time error if a statement cannot be executed because it is unreachable. The question is now, why is line 20 unreachable? If it is because of the assert then surely line 6 would also be unreachable. The answer must be something other than assert.

Examine the following:

A while statement can complete normally if and only if at least one of the following is true:

- The while statement is reachable and the condition expression is not a constant expression with value true.

-There is a reachable break statement that exits the while statement.


19. Which cannot directly cause a thread to stop executing?



Explanation:

notify() - wakes up a single thread that is waiting on this object's monitor.


20.
Which two are valid constructors for Thread?

Thread(Runnable r, String name)
Thread()
Thread(int priority)
Thread(Runnable r, ThreadGroup g)
Thread(Runnable r, int priority)



Explanation:

(1) and (2) are both valid constructors for Thread.

(3), (4), and (5) are not legal Thread constructors, although (4) is close. If you reverse the arguments in (4), you'd have a valid constructor.


21.
What will be the output of the program?

public class SwitchTest 
{  
    public static void main(String[] args) 
    {
        System.out.println("value =" + switchIt(4)); 
    } 
    public static int switchIt(int x) 
    {
        int j = 1;  
        switch (x) 
        { 
            case l: j++; 
            case 2: j++;  
            case 3: j++; 
            case 4: j++; 
            case 5: j++; 
            default: j++; 
            } 
        return j + x;  
    } 
}



Explanation:

Because there are no break statements, once the desired result is found, the program continues though each of the remaining options.


22.
class Boo 
{
    Boo(String s) { }
    Boo() { }
}
class Bar extends Boo 
{
    Bar() { }
    Bar(String s) {super(s);}
    void zoo() 
    {
    // insert code here
    }
}
which one create an anonymous inner class from within class Bar?



Explanation:

Option B is correct because anonymous inner classes are no different from any other class when it comes to polymorphism. That means you are always allowed to declare a reference variable of the superclass type and have that reference variable refer to an instance of a subclass type, which in this case is an anonymous subclass of Bar. Since Bar is a subclass of Boo, it all works.

Option A is incorrect because it passes an int to the Boo constructor, and there is no matching constructor in the Boo class.

Option C is incorrect because it violates the rules of polymorphism—you cannot refer to a superclass type using a reference variable declared as the subclass type. The superclass is not guaranteed to have everything the subclass has.

Option D uses incorrect syntax.


23. Which is true about a method-local inner class?



Explanation:

Option B is correct because a method-local inner class can be abstract, although it means a subclass of the inner class must be created if the abstract class is to be used (so an abstract method-local inner class is probably not useful).

Option A is incorrect because a method-local inner class does not have to be declared final (although it is legal to do so).

C and D are incorrect because a method-local inner class cannot be made public (remember-you cannot mark any local variables as public), or static.


24.
public class MyRunnable implements Runnable 
{
    public void run() 
    {
        // some code here
    }
}
which of these will create and start this thread?



Explanation:

Because the class implements Runnable, an instance of it has to be passed to the Thread constructor, and then the instance of the Thread has to be started.

A is incorrect. There is no constructor like this for Runnable because Runnable is an interface, and it is illegal to pass a class or interface name to any constructor.

B is incorrect for the same reason; you can't pass a class or interface name to any constructor.

D is incorrect because MyRunnable doesn't have a start() method, and the only start() method that can start a thread of execution is the start() in the Thread class.


25.
What will be the output of the program?

class Test 
{
    public static void main(String [] args) 
    {
        int x=20;
        String sup = (x < 15) ? "small" : (x < 22)? "tiny" : "huge";
        System.out.println(sup);
    }
}



Explanation:

This is an example of a nested ternary operator. The second evaluation (x < 22) is true, so the "tiny" value is assigned to sup.


26.
What will be the output of the program?

class Test 
{
    static int s;
    public static void main(String [] args) 
    {
        Test p = new Test();
        p.start();
        System.out.println(s);
    }

    void start() 
    {
        int x = 7;
        twice(x);
        System.out.print(x + " ");
    }

    void twice(int x) 
    {
        x = x*2;
        s = x;
    }
}



Explanation:

The int x in the twice() method is not the same int x as in the start() method. Start()'s x is not affected by the twice() method. The instance variable s is updated by twice()'s x, which is 14.


27.
What will be the output of the program?

class BoolArray 
{
    boolean [] b = new boolean[3];
    int count = 0;

    void set(boolean [] x, int i) 
    {
        x[i] = true;
        ++count;
    }

    public static void main(String [] args) 
    {
        BoolArray ba = new BoolArray();
        ba.set(ba.b, 0);
        ba.set(ba.b, 2);
        ba.test();
    }

    void test() 
    {
        if ( b[0] && b[1] | b[2] )
            count++;
        if ( b[1] && b[(++count - 2)] )
            count += 7;
        System.out.println("count = " + count);
    }
}



Explanation:

The reference variables b and x both refer to the same boolean array. count is incremented for each call to the set() method, and once again when the first if test is true. Because of the && short circuit operator, count is not incremented during the second if test.


28.
Which three are valid declarations of a char?

char c1 = 064770;
char c2 = 'face';
char c3 = 0xbeef;
char c4 = u0022;
char c5 = 'iface';
char c6 = 'uface';



Explanation:

(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'.


29. Which interface does java.util.Hashtable implement?



Explanation:

Hash table based implementation of the Map interface.


30.
Which three statements are true?

Assertion checking is typically enabled when a program is deployed.
It is never appropriate to write code to handle failure of an assert statement.
Assertion checking is typically enabled during program development and testing.
Assertion checking can be selectively enabled or disabled on a per-package basis, but not on a per-class basis.
Assertion checking can be selectively enabled or disabled on both a per-package basis and a per-class basis.



Explanation:

(1) is wrong. It's just not true.

(2) is correct. You're never supposed to handle an assertion failure.

(3) is correct. Assertions let you test your assumptions during development, but the assertion code—in effect—evaporates when the program is deployed, leaving behind no overhead or debugging code to track down and remove.

(4) is wrong. See the explanation for (5) below.

(5) is correct. Assertion checking can be selectively enabled or disabled on a per-package basis. Note that the package default assertion status determines the assertion status for classes initialized in the future that belong to the named package or any of its "subpackages".

The assertion status can be set for a named top-level class and any nested classes contained therein. This setting takes precedence over the class loader's default assertion status, and over any applicable per-package default. If the named class is not a top-level class, the change of status will have no effect on the actual assertion status of any class




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