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Java Questions And Answers Sample Test 6


Sample Java Test 6 for you to Practice. Evaluate your Java Test 6 test answering skills by trying the online Java Sample Test 6 and know your score.

Java Test 6


You have scored 3 out of 10
You Scored:2
Total Questions:2
Attended Questions:2
Correct Answered:2
1. What is the name of the method used to start a thread execution?



Explanation:

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

Option C 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 D is wrong. The resume() method is deprecated. It resumes a suspended thread.


2.
void start() {  
    A a = new A(); 
    B b = new B(); 
    a.s(b);  
    b = null; /* Line 5 */
    a = null;  /* Line 6 */
    System.out.println("start completed"); /* Line 7 */
When is the B object, created in line 3, eligible for garbage collection?





3.
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.

The while statement at line 11 is infinite and there is no break statement therefore line 14 is unreachable. You can test this with the following code:

public class Test80
{
public void foo()
{
assert false;
assert false;
}
public void bar()
{
while(true)
{
assert false;
break;
}
assert false;
}
}


4.
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.


5. Which is a valid keyword in java?



Explanation:

interface is a valid keyword.

Option B is wrong because although "String" is a class type in Java, "string" is not a keyword.

Option C is wrong because "Float" is a class type. The keyword for the Java primitive is float.

Option D is wrong because "unsigned" is a keyword in C/C++ but not in Java.


6. Which of the following will not directly cause a thread to stop?



Explanation:

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

Option B is wrong. wait() causes the current thread to wait until another thread invokes the notify() method or the notifyAll() method for this object.

Option C is wrong. Methods of the InputStream class block until input data is available, the end of the stream is detected, or an exception is thrown. Blocking means that a thread may stop until certain conditions are met.

Option D is wrong. sleep() - Causes the currently executing thread to sleep (temporarily cease execution) for a specified number of milliseconds. The thread does not lose ownership of any monitors.


7.
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.


8.
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.


9.
What will be the output of the program?

public class MyProgram 
{
    public static void main(String args[])
    {
        try 
        {
            System.out.print("Hello world ");
        }
        finally 
        {
            System.out.println("Finally executing ");
        }
    }
}



Explanation:

Finally clauses are always executed. The program will first execute the try block, printing Hello world, and will then execute the finally block, printing Finally executing.

Option A, B, and C are incorrect based on the program logic described above. Remember that either a catch or a finally statement must follow a try. Since the finally is present, the catch is not required.


10.
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.


11. 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.


12. Which of the following will directly stop the execution of a Thread?



Explanation:

Option A is correct. wait() causes the current thread to wait until another thread invokes the notify() method or the notifyAll() method for this object.

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

Option C is wrong. notifyAll() - wakes up all threads that are waiting on this object's monitor.

Option D is wrong. Typically, releasing a lock means the thread holding the lock (in other words, the thread currently in the synchronized method) exits the synchronized method. At that point, the lock is free until some other thread enters a synchronized method on that object. Does entering/exiting synchronized code mean that the thread execution stops? Not necessarily because the thread can still run code that is not synchronized. I think the word directly in the question gives us a clue. Exiting synchronized code does not directly stop the execution of a thread.


13. Which is true about an anonymous inner class?



Explanation:

Option C is correct because the syntax of an anonymous inner class allows for only one named type after the new, and that type must be either a single interface (in which case the anonymous class implements that one interface) or a single class (in which case the anonymous class extends that one class).

Option A, B, D, and E are all incorrect because they don't follow the syntax rules described in the response for answer Option C.


14.
Which three guarantee that a thread will leave the running state?

yield()
wait()
notify()
notifyAll()
sleep(1000)
aLiveThread.join()
Thread.killThread()



Explanation:

(2) is correct because wait() always causes the current thread to go into the object's wait pool.

(5) is correct because sleep() will always pause the currently running thread for at least the duration specified in the sleep argument (unless an interrupted exception is thrown).

(6) is correct because, assuming that the thread you're calling join() on is alive, the thread calling join() will immediately block until the thread you're calling join() on is no longer alive.

(1) is wrong, but tempting. The yield() method is not guaranteed to cause a thread to leave the running state, although if there are runnable threads of the same priority as the currently running thread, then the current thread will probably leave the running state.

(3) and (4) are incorrect because they don't cause the thread invoking them to leave the running state.

(7) is wrong because there's no such method.


15.
class X implements Runnable 
    public static void main(String args[]) 
    {
        /* Missing code? */
    } 
    public void run() {} 
}
Which of the following line of code is suitable to start a thread ?





16. 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.


17.
What is the numerical range of a char?



Explanation:

A char is really a 16-bit integer behind the scenes, so it supports 216 (from 0 to 65535) values.


18. Which method must be defined by a class implementing the java.lang.Runnable interface?



Explanation:

Option B is correct because in an interface all methods are abstract by default therefore they must be overridden by the implementing class. The Runnable interface only contains 1 method, the void run() method therefore it must be implemented.

Option A and D are incorrect because they are narrowing the access privileges i.e. package(default) access is narrower than public access.

Option C is not method in the Runnable interface therefore it is incorrect.


19.
What will be the output of the program?

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

    void start() 
    {
        String s1 = "slip";
        String s2 = fix(s1);
        System.out.println(s1 + " " + s2);
    }

    String fix(String s1) 
    {
        s1 = s1 + "stream";
        System.out.print(s1 + " ");
        return "stream";
    }
}



Explanation:

When the fix() method is first entered, start()'s s1 and fix()'s s1 reference variables both refer to the same String object (with a value of "slip"). Fix()'s s1 is reassigned to a new object that is created when the concatenation occurs (this second String object has a value of "slipstream"). When the program returns to start(), another String object is created, referred to by s2 and with a value of "stream".


20.
class Test 
{  
    private Demo d; 
    void start() 
    {  
        d = new Demo(); 
        this.takeDemo(d); /* Line 7 */
    } /* Line 8 */
    void takeDemo(Demo demo) 
    { 
        demo = null;  
        demo = new Demo(); 
    } 
}
When is the Demo object eligible for garbage collection?



Explanation:

Option D is correct. By a process of elimination.

Option A is wrong. The variable d is a member of the Test class and is never directly set to null.

Option B is wrong. A copy of the variable d is set to null and not the actual variable d.

Option C is wrong. The variable d exists outside the start() method (it is a class member). So, when the start() method finishes the variable d still holds a reference.


21. What is the numerical range of char?



Explanation:

The char type is integral but unsigned. The range of a variable of type char is from 0 to 216-1 or 0 to 65535. Java characters are Unicode, which is a 16-bit encoding capable of representing a wide range of international characters. If the most significant nine bits of a char are 0, then the encoding is the same as seven-bit ASCII.


22.
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.


23.
What will be the output of the program?

public class X 
{  
    public static void main(String [] args) 
    {
        try 
        {
            badMethod();  
            System.out.print("A");  
        } 
        catch (RuntimeException ex) /* Line 10 */
        { 
            System.out.print("B"); 
        } 
        catch (Exception ex1) 
        { 
            System.out.print("C"); 
        } 
        finally 
        {
            System.out.print("D"); 
        } 
        System.out.print("E"); 
    } 
    public static void badMethod() 
    { 
        throw new RuntimeException(); 
    } 
}



Explanation:

A Run time exception is thrown and caught in the catch statement on line 10. All the code after the finally statement is run because the exception has been caught.


24. 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.


25.
Which four can be thrown using the throw statement?

Error
Event
Object
Throwable
Exception
RuntimeException



Explanation:

The (1), (4), (5) and (6) are the only four that can be thrown.

An Error is a subclass of Throwable that indicates serious problems that a reasonable application should not try to catch.

The Throwable class is the superclass of all errors and exceptions in the Java language.

The class Exception and its subclasses are a form of Throwable that indicates conditions that a reasonable application might want to catch (checked exceptions)

RuntimeException is the superclass of those exceptions that can be thrown during the normal operation of the Java Virtual Machine.


26.
What will be the output of the program?

class Equals 
{
    public static void main(String [] args) 
    {
        int x = 100;
        double y = 100.1;
        boolean b = (x = y); /* Line 7 */
        System.out.println(b);
    }
}



Explanation:

The code will not compile because in line 7, the line will work only if we use (x==y) in the line. The == operator compares values to produce a boolean, whereas the = operator assigns a value to variables.

Option A, B, and D are incorrect because the code does not get as far as compiling. If we corrected this code, the output would be false.


27.
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.


28.
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.


29. Which is valid declaration of a float?



Explanation:

Option A is valid declaration of float.

Option B is incorrect because any literal number with a decimal point u declare the computer will implicitly cast to double unless you include "F or f"

Option C is incorrect because it is a String.

Option D is incorrect because "d" tells the computer it is a double so therefore you are trying to put a double value into a float variable i.e there might be a loss of precision.


30.
What will be the output of the program?

int i = 1, j = 10; 
do 
{
    if(i > j) 
    {
        break; 
    } 
    j--; 
} while (++i < 5); 
System.out.println("i = " + i + " and j = " + j);



Explanation:

This loop is a do-while loop, which always executes the code block within the block at least once, due to the testing condition being at the end of the loop, rather than at the beginning. This particular loop is exited prematurely if i becomes greater than j.

The order is, test i against j, if bigger, it breaks from the loop, decrements j by one, and then tests the loop condition, where a pre-incremented by one i is tested for being lower than 5. The test is at the end of the loop, so i can reach the value of 5 before it fails. So it goes, start:

1, 10

2, 9

3, 8

4, 7

5, 6 loop condition fails.




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