Online Java Programming Test - Java Programming Test 4
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- Total number of questions: 20.
- Time allotted: 30 minutes.
- Each question carries 1 mark; there are no negative marks.
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- All the best!
Marks : 2/20
Test Review : View answers and explanation for this test.
interface Base
{
boolean m1 ();
byte m2(short s);
}
- interface Base2 implements Base {}
-
abstract class Class2 extends Base
{ public boolean m1(){ return true; }} - abstract class Class2 implements Base {}
- abstract class Class2 implements Base
{ public boolean m1(){ return (7 > 4); }} - abstract class Class2 implements Base
{ protected boolean m1(){ return (5 > 7) }}
(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.
Access modifiers dictate which classes, not which instances, may access features.
Methods and variables are collectively known as members. Method and variable members are given access control in exactly the same way.
private makes a member accessible only from within its own class
protected makes a member accessible only to classes in the same package or subclass of the class
default access is very similar to protected (make sure you spot the difference) default access makes a member accessible only to classes in the same package.
public means that all other classes regardless of the package that they belong to, can access the member (assuming the class itself is visible)
final makes it impossible to extend a class, when applied to a method it prevents a method from being overridden in a subclass, when applied to a variable it makes it impossible to reinitialise a variable once it has been initialised
abstract declares a method that has not been implemented.
transient indicates that a variable is not part of the persistent state of an object.
volatile indicates that a thread must reconcile its working copy of the field with the master copy every time it accesses the variable.
After examining the above it should be obvious that the access modifier that provides the most restrictions for methods to be accessed from the subclasses of the class from another package is C - protected. A is also a contender but C is more restrictive, B would be the answer if the constraint was the "same package" instead of "any package" in other words the subclasses clause in the question eliminates default.
class A
{
protected int method1(int a, int b)
{
return 0;
}
}
Option A is correct - because the class that extends A is just simply overriding method1.
Option B is wrong - because it can't override as there are less access privileges in the subclass method1.
Option C is wrong - because to override it, the return type needs to be an integer. The different return type means that the method is not overriding but the same argument list means that the method is not overloading. Conflict - compile time error.
Option D is wrong - because you can't override a method and make it a class method i.e. using static.
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;
}
}
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.
class Test
{
public static void main(String [] args)
{
Test p = new Test();
p.start();
}
void start()
{
boolean b1 = false;
boolean b2 = fix(b1);
System.out.println(b1 + " " + b2);
}
boolean fix(boolean b1)
{
b1 = true;
return b1;
}
}
The boolean b1 in the fix() method is a different boolean than the b1 in the start() method. The b1 in the start() method is not updated by the fix() method.
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";
}
}
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".
import java.awt.*;
class Ticker extends Component
{
public static void main (String [] args)
{
Ticker t = new Ticker();
/* Missing Statements ? */
}
}
- boolean test = (Component instanceof t);
- boolean test = (t instanceof Ticker);
- boolean test = t.instanceof(Ticker);
- boolean test = (t instanceof Component);
(2) is correct because class type Ticker is part of the class hierarchy of t; therefore it is a legal use of the instanceof operator. (4) is also correct because Component is part of the hierarchy of t, because Ticker extends Component.
(1) is incorrect because the syntax is wrong. A variable (or null) always appears before the instanceof operator, and a type appears after it. (3) is incorrect because the statement is used as a method (t.instanceof(Ticker);), which is illegal.
public void test(int x)
{
int odd = 1;
if(odd) /* Line 4 */
{
System.out.println("odd");
}
else
{
System.out.println("even");
}
}
The compiler will complain because of incompatible types (line 4), the if expects a boolean but it gets an integer.
public class Test
{
public static void main(String args[])
{
int i = 1, j = 0;
switch(i)
{
case 2: j += 6;
case 4: j += 1;
default: j += 2;
case 0: j += 4;
}
System.out.println("j = " + j);
}
}
Because there are no break statements, the program gets to the default case and adds 2 to j, then goes to case 0 and adds 4 to the new j. The result is j = 6.
int i = 1, j = 10;
do
{
if(i > j)
{
break;
}
j--;
} while (++i < 5);
System.out.println("i = " + i + " and j = " + j);
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.
import java.io.*;
public class MyProgram
{
public static void main(String args[])
{
FileOutputStream out = null;
try
{
out = new FileOutputStream("test.txt");
out.write(122);
}
catch(IOException io)
{
System.out.println("IO Error.");
}
finally
{
out.close();
}
}
}
Any method (in this case, the main() method) that throws a checked exception (in this case, out.close() ) must be called within a try clause, or the method must declare that it throws the exception. Either main() must declare that it throws an exception, or the call to out.close() in the finally block must fall inside a (in this case nested) try-catch block.
Option A is correct. If the class specified in the catch clause does have subclasses, any exception object that subclasses the specified class will be caught as well.
Option B is wrong. The error class is a subclass of Throwable and not Runtime Exception.
Option C is wrong. You do not catch this class of error.
Option D is wrong. An exception can be thrown to the next method higher up the call stack.
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.
import java.util.*;
class H
{
public static void main (String[] args)
{
Object x = new Vector().elements();
System.out.print((x instanceof Enumeration)+",");
System.out.print((x instanceof Iterator)+",");
System.out.print(x instanceof ListIterator);
}
}
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.
TreeSet map = new TreeSet();
map.add("one");
map.add("two");
map.add("three");
map.add("four");
map.add("one");
Iterator it = map.iterator();
while (it.hasNext() )
{
System.out.print( it.next() + " " );
}
TreeSet assures no duplicate entries; also, when it is accessed it will return elements in natural order, which typically means alphabetical.
class Foo
{
class Bar{ }
}
class Test
{
public static void main (String [] args)
{
Foo f = new Foo();
/* Line 10: Missing statement ? */
}
}
Option B is correct because the syntax is correct-using both names (the enclosing class and the inner class) in the reference declaration, then using a reference to the enclosing class to invoke new on the inner class.
Option A, C and D all use incorrect syntax. A is incorrect because it doesn't use a reference to the enclosing class, and also because it includes both names in the new.
C is incorrect because it doesn't use the enclosing class name in the reference variable declaration, and because the new syntax is wrong.
D is incorrect because it doesn't use the enclosing class name in the reference variable declaration.
Option A is Correct. The run() method to 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. The start() method causes this thread to begin execution; the Java Virtual Machine calls the run method of this thread.
Option C is wrong. The stop() method is deprecated. It forces the thread to stop executing.
Option D is wrong. Is the main entry point for an application.
- yield()
- wait()
- notify()
- notifyAll()
- sleep(1000)
- aLiveThread.join()
- Thread.killThread()
(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.
public class Test
{
public static void main(String[] args)
{
int x = 0;
assert (x > 0) ? "assertion failed" : "assertion passed" ;
System.out.println("finished");
}
}
Compilation Fails. You can't use the Assert statement in a similar way to the ternary operator. Don't confuse.
public class BoolTest
{
public static void main(String [] args)
{
int result = 0;
Boolean b1 = new Boolean("TRUE");
Boolean b2 = new Boolean("true");
Boolean b3 = new Boolean("tRuE");
Boolean b4 = new Boolean("false");
if (b1 == b2) /* Line 10 */
result = 1;
if (b1.equals(b2) ) /* Line 12 */
result = result + 10;
if (b2 == b4) /* Line 14 */
result = result + 100;
if (b2.equals(b4) ) /* Line 16 */
result = result + 1000;
if (b2.equals(b3) ) /* Line 18 */
result = result + 10000;
System.out.println("result = " + result);
}
}
Line 10 fails because b1 and b2 are two different objects. Lines 12 and 18 succeed because the Boolean String constructors are case insensitive. Lines 14 and 16 fail because true is not equal to false.