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.
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.
public class ReturnIt
{
returnType methodA(byte x, double y) /* Line 3 */
{
return (long)x / y * 2;
}
}
However A, B and C are all wrong. Each of these would result in a narrowing conversion. Whereas we want a widening conversion, therefore the only correct answer is D. Don't be put off by the long cast, this applies only to the variable x and not the rest of the expression. It is the variable y (of type double) that forces the widening conversion to double.
Java's widening conversions are:
- From a byte to a short, an int, a long, a float, or a double.
- From a short, an int, a long, a float, or a double.
- From a char to an int, a long, a float, or a double.
- From an int to a long, a float, or a double.
- From a long to a float, or a double.
- From a float to a double.
interface DoMath
{
double getArea(int rad);
}
interface MathPlus
{
double getVol(int b, int h);
}
/* Missing Statements ? */
- class AllMath extends DoMath { double getArea(int r); }
- interface AllMath implements MathPlus { double getVol(int x, int y); }
- interface AllMath extends DoMath { float getAvg(int h, int l); }
- class AllMath implements MathPlus { double getArea(int rad); }
- abstract class AllMath implements DoMath, MathPlus { public double getArea(int rad) { return rad * rad * 3.14; } }
(3) are (5) are correct because interfaces and abstract classes do not need to fully implement the interfaces they extend or implement (respectively).
(1) is incorrect because a class cannot extend an interface. (2) is incorrect because an interface cannot implement anything. (4) is incorrect because the method being implemented is from the wrong interface.
package testpkg.p1;
public class ParentUtil
{
public int x = 420;
protected int doStuff() { return x; }
}
package testpkg.p2;
import testpkg.p1.ParentUtil;
public class ChildUtil extends ParentUtil
{
public static void main(String [] args)
{
new ChildUtil().callStuff();
}
void callStuff()
{
System.out.print("this " + this.doStuff() ); /* Line 18 */
ParentUtil p = new ParentUtil();
System.out.print(" parent " + p.doStuff() ); /* Line 20 */
}
}
The ParentUtil instance p cannot be used to access the doStuff() method. Because doStuff() has protected access, and the ChildUtil class is not in the same package as the ParentUtil class, doStuff() can be accessed only by instances of the ChildUtil class (a subclass of ParentUtil).
Option A, B and D are incorrect because of the access rules described previously.
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;
}
}
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
- 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).
public class MyProgram
{
public static void main(String args[])
{
try
{
System.out.print("Hello world ");
}
finally
{
System.out.println("Finally executing ");
}
}
}
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.
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.
public class Test
{
public static void main (String args[])
{
String str = NULL;
System.out.println(str);
}
}
Option B is correct because to set the value of a String variable to null you must use "null" and not "NULL".
public class Test
{
private static float[] f = new float[2];
public static void main (String[] args)
{
System.out.println("f[0] = " + f[0]);
}
}
The choices are between Option A and B, what this question is really testing is your knowledge of default values of an initialized array. This is an array type float i.e. it is a type that uses decimal point numbers therefore its initial value will be 0.0 and not 0
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.
Option B is correct because a static nested class is not tied to an instance of the enclosing class, and thus can't access the nonstatic members of the class (just as a static method can't access nonstatic members of a class).
Option A is incorrect because static nested classes do not need (and can't use) a reference to an instance of the enclosing class.
Option C is incorrect because static nested classes can declare and define nonstatic members.
Option D is wrong because it just is. There's no rule that says an inner or nested class has to extend anything.
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.
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.
class Test116
{
static final StringBuffer sb1 = new StringBuffer();
static final StringBuffer sb2 = new StringBuffer();
public static void main(String args[])
{
new Thread()
{
public void run()
{
synchronized(sb1)
{
sb1.append("A");
sb2.append("B");
}
}
}.start();
new Thread()
{
public void run()
{
synchronized(sb1)
{
sb1.append("C");
sb2.append("D");
}
}
}.start(); /* Line 28 */
System.out.println (sb1 + " " + sb2);
}
}
Can you guarantee the order in which threads are going to run? No you can't. So how do you know what the output will be? The output cannot be determined.
add this code after line 28:
try { Thread.sleep(5000); } catch(InterruptedException e) { }
and you have some chance of predicting the outcome.
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
Option D is correct.
Option C is wrong. See the note above on Islands of Isolation (An object is eligible for garbage collection when no live thread can access it - even though there might be references to it).
Option B is wrong. "Never again be used" does not mean that there are no more references to the object.
Option A is wrong. Even though Java applications can run out of memory there another answer supplied that is more right.
The return value of the Math.abs() method is always the same as the type of the parameter passed into that method.
In the case of A, an integer is passed in and so the result is also an integer which is fine for assignment to "int a".
The values used in B, C & D respectively are a double, a float and a long. The compiler will complain about a possible loss of precision if we try to assign the results to an "int".
public class Myfile
{
public static void main (String[] args)
{
String biz = args[1];
String baz = args[2];
String rip = args[3];
System.out.println("Arg is " + rip);
}
}
Arguments start at array element 0 so the fourth arguement must be 2 to produce the correct output.
public class SqrtExample
{
public static void main(String [] args)
{
double value = -9.0;
System.out.println( Math.sqrt(value));
}
}
The sqrt() method returns NaN (not a number) when it's argument is less than zero.