Online Java Programming Test - Java Programming Test - Random
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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.
public interface Foo
{
int k = 4; /* Line 3 */
}
- final int k = 4;
- public int k = 4;
- static int k = 4;
- abstract int k = 4;
- volatile int k = 4;
- protected int k = 4;
(1), (2) and (3) are correct. Interfaces can have constants, which are always implicitly public, static, and final. Interface constant declarations of public, static, and final are optional in any combination.
- int -> 0
- String -> "null"
- Dog -> null
- char -> '\u0000'
- float -> 0.0f
- boolean -> true
(1), (3), (4), (5) are the correct statements.
(2) is wrong because the default value for a String (and any other object reference) is null, with no quotes.
(6) is wrong because the default value for boolean elements is false.
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.
The only two real contenders are C and D. Protected access Option C makes a member accessible only to classes in the same package or subclass of the class. While default access Option D makes a member accessible only to classes in the same package.
class A
{
final public int GetResult(int a, int b) { return 0; }
}
class B extends A
{
public int GetResult(int a, int b) {return 1; }
}
public class Test
{
public static void main(String args[])
{
B b = new B();
System.out.println("x = " + b.GetResult(0, 1));
}
}
The code doesn't compile because the method GetResult() in class A is final and so cannot be overridden.
class Test
{
public static void main(String [] args)
{
int x= 0;
int y= 0;
for (int z = 0; z < 5; z++)
{
if (( ++x > 2 ) && (++y > 2))
{
x++;
}
}
System.out.println(x + " " + y);
}
}
In the first two iterations x is incremented once and y is not because of the short circuit && operator. In the third and forth iterations x and y are each incremented, and in the fifth iteration x is doubly incremented and y is incremented.
class Two
{
byte x;
}
class PassO
{
public static void main(String [] args)
{
PassO p = new PassO();
p.start();
}
void start()
{
Two t = new Two();
System.out.print(t.x + " ");
Two t2 = fix(t);
System.out.println(t.x + " " + t2.x);
}
Two fix(Two tt)
{
tt.x = 42;
return tt;
}
}
In the fix() method, the reference variable tt refers to the same object (class Two) as the t reference variable. Updating tt.x in the fix() method updates t.x (they are one in the same object). Remember also that the instance variable x in the Two class is initialized to 0.
- 32/4
- (8 >> 2) << 4
- 2^5
- 128 >>> 2
- 2 >> 5
(2) and (4) are correct. (2) and (4) both evaluate to 32. (2) is shifting bits right then left using the signed bit shifters >> and <<. (4) is shifting bits using the unsigned operator >>>, but since the beginning number is positive the sign is maintained.
(1) evaluates to 8, (3) looks like 2 to the 5th power, but ^ is the Exclusive OR operator so (3) evaluates to 7. (5) evaluates to 0 (2 >> 5 is not 2 to the 5th).
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");
}
}
}
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.
/* Missing Statement ? */
public class foo
{
public static void main(String[]args)throws Exception
{
java.io.PrintWriter out = new java.io.PrintWriter();
new java.io.OutputStreamWriter(System.out,true);
out.println("Hello");
}
}
The usual method for using/importing the java packages/classes is by using an import statement at the top of your code. However it is possible to explicitly import the specific class that you want to use as you use it which is shown in the code above. The disadvantage of this however is that every time you create a new object you will have to use the class path in the case "java.io" then the class name in the long run leading to a lot more typing.
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.
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.
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.
public class Q126 implements Runnable
{
private int x;
private int y;
public static void main(String [] args)
{
Q126 that = new Q126();
(new Thread(that)).start( ); /* Line 8 */
(new Thread(that)).start( ); /* Line 9 */
}
public synchronized void run( ) /* Line 11 */
{
for (;;) /* Line 13 */
{
x++;
y++;
System.out.println("x = " + x + "y = " + y);
}
}
}
The synchronized code is the key to answering this question. Because x and y are both incremented inside the synchronized method they are always incremented together. Also keep in mind that the two threads share the same reference to the Q126 object.
Also note that because of the infinite loop at line 13, only one thread ever gets to execute.
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;
}
}
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.
Ref: http://www.javaworld.com/javaworld/javaqa/2000-05/03-qa-0526-pass.html
class Bar { }
class Test
{
Bar doBar()
{
Bar b = new Bar(); /* Line 6 */
return b; /* Line 7 */
}
public static void main (String args[])
{
Test t = new Test(); /* Line 11 */
Bar newBar = t.doBar(); /* Line 12 */
System.out.println("newBar");
newBar = new Bar(); /* Line 14 */
System.out.println("finishing"); /* Line 15 */
}
}
Option B is correct. All references to the Bar object created on line 6 are destroyed when a new reference to a new Bar object is assigned to the variable newBar on line 14. Therefore the Bar object, created on line 6, is eligible for garbage collection after line 14.
Option A is wrong. This actually protects the object from garbage collection.
Option C is wrong. Because the reference in the doBar() method is returned on line 7 and is stored in newBar on line 12. This preserver the object created on line 6.
Option D is wrong. Not applicable because the object is eligible for garbage collection after line 14.
Option D is correct. When an object is no longer referenced, it may be reclaimed by the garbage collector. If an object declares a finalizer, the finalizer is executed before the object is reclaimed to give the object a last chance to clean up resources that would not otherwise be released. When a class is no longer needed, it may be unloaded.
Option A is wrong. I found 4 delete() methods in all of the Java class structure. They are:
- delete() - Method in class java.io.File : Deletes the file or directory denoted by this abstract pathname.
- delete(int, int) - Method in class java.lang.StringBuffer : Removes the characters in a substring of this StringBuffer.
- delete(int, int) - Method in interface javax.accessibility.AccessibleEditableText : Deletes the text between two indices
- delete(int, int) - Method in class : javax.swing.text.JTextComponent.AccessibleJTextComponent; Deletes the text between two indices
None of these destroy the object to which they belong.
Option B is wrong. I found 19 finalize() methods. The most interesting, from this questions point of view, was the finalize() method in class java.lang.Object which is called by the garbage collector on an object when garbage collection determines that there are no more references to the object. This method does not destroy the object to which it belongs.
Option C is wrong. But it is interesting. The Runtime class has many methods, two of which are:
- getRuntime() - Returns the runtime object associated with the current Java application.
- gc() - Runs the garbage collector. Calling this method suggests that the Java virtual machine expend effort toward recycling unused objects in order to make the memory they currently occupy available for quick reuse. When control returns from the method call, the virtual machine has made its best effort to recycle all discarded objects. Interesting as this is, it doesn't destroy the object.
Option B is correct. If an object can be accessed from a live thread, it can't be garbage collected.
Option A is wrong. Runtime.gc() asks the garbage collector to run, but the garbage collector never makes any guarantees about when it will run or what unreachable objects it will free from memory.
Option C is wrong. The garbage collector runs immediately the system is out of memory before an OutOfMemoryException is thrown by the JVM.
Option D is wrong. If this were the case then the garbage collector would actively hang onto objects until a program finishes - this goes against the purpose of the garbage collector.
public class ObjComp
{
public static void main(String [] args )
{
int result = 0;
ObjComp oc = new ObjComp();
Object o = oc;
if (o == oc)
result = 1;
if (o != oc)
result = result + 10;
if (o.equals(oc) )
result = result + 100;
if (oc.equals(o) )
result = result + 1000;
System.out.println("result = " + result);
}
}
Even though o and oc are reference variables of different types, they are both referring to the same object. This means that == will resolve to true and that the default equals() method will also resolve to true.
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 */
}
}
This question is about the + (plus) operator and the overriden + (string cocatanation) operator. The rules that apply when you have a mixed expression of numbers and strings are:
If either operand is a String, the + operator concatenates the operands.
If both operands are numeric, the + operator adds the operands.
The expression on line 6 above can be read as "Add the values i1 and i2 together, then take the sum and convert it to a string and concatenate it with the String from the variable s1". In code, the compiler probably interprets the expression on line 8 above as:
System.out.println( new StringBuffer()
.append(new Integer(i1 + i2).toString())
.append(s1)
.toString() );