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.
- DO NOT refresh the page.
- All the best!
Marks : 2/20
Test Review : View answers and explanation for this test.
Option A sets the String reference to null.
Option B is wrong because null cannot be in single quotes.
Option C is wrong because there are multiple characters between the single quotes ('abc').
Option D is wrong because you can't cast a char (primitive) to a String (object).
default access is the "package oriented" access modifier.
Option A and C are wrong because public and protected are less restrictive. Option B and D are wrong because abstract and synchronized are not access modifiers.
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.
- private int getArea();
- public float getVol(float x);
- public void main(String [] args);
- public static void main(String [] args);
- boolean setFlag(Boolean [] test);
(2), (3), and (5). These are all valid interface method signatures.
(1), is incorrect because an interface method must be public; if it is not explicitly declared public it will be made public implicitly. (4) is incorrect because interface methods cannot be static.
public class ArrayTest
{
public static void main(String[ ] args)
{
float f1[ ], f2[ ];
f1 = new float[10];
f2 = f1;
System.out.println("f2[0] = " + f2[0]);
}
}
Option A is correct. When you create an array (f1 = new float[10];) the elements are initialises to the default values for the primitive data type (float in this case - 0.0), so f1 will contain 10 elements each with a value of 0.0. f2 has been declared but has not been initialised, it has the ability to reference or point to an array but as yet does not point to any array. f2 = f1; copies the reference (pointer/memory address) of f1 into f2 so now f2 points at the array pointed to by f1.
This means that the values returned by f2 are the values returned by f1. Changes to f1 are also changes to f2 because both f1 and f2 point to the same array.
- The default constructor initialises method variables.
- The default constructor has the same access as its class.
- The default constructor invokes the no-arg constructor of the superclass.
- If a class lacks a no-arg constructor, the compiler always creates a default constructor.
- The compiler creates a default constructor only when there are no other constructors for the class.
(2) sounds correct as in the example below
class CoffeeCup {
private int innerCoffee;
public CoffeeCup() {
}
public void add(int amount) {
innerCoffee += amount;
}
//...
}
(3) is correct. The Java compiler generates at least one instance initialisation method for every class it compiles. In the Java class file, the instance initialisation method is named "<init>." For each constructor in the source code of a class, the Java compiler generates one <init>() method. If the class declares no constructors explicitly, the compiler generates a default no-arg constructor that just invokes the superclass's no-arg constructor. As with any other constructor, the compiler creates an <init>() method in the class file that corresponds to this default constructor.
(5) is correct. The compiler creates a default constructor if you do not declare any constructors in your class.
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.
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 SC2
{
public static void main(String [] args)
{
SC2 s = new SC2();
s.start();
}
void start()
{
int a = 3;
int b = 4;
System.out.print(" " + 7 + 2 + " ");
System.out.print(a + b);
System.out.print(" " + a + b + " ");
System.out.print(foo() + a + b + " ");
System.out.println(a + b + foo());
}
String foo()
{
return "foo";
}
}
Because all of these expressions use the + operator, there is no precedence to worry about and all of the expressions will be evaluated from left to right. If either operand being evaluated is a String, the + operator will concatenate the two operands; if both operands are numeric, the + operator will add the two operands.
int x = 3;
int y = 1;
if (x = y) /* Line 3 */
{
System.out.println("x =" + x);
}
Line 3 uses an assignment as opposed to comparison. Because of this, the if statement receives an integer value instead of a boolean. And so the compilation fails.
boolean bool = true;
if(bool = false) /* Line 2 */
{
System.out.println("a");
}
else if(bool) /* Line 6 */
{
System.out.println("b");
}
else if(!bool) /* Line 10 */
{
System.out.println("c"); /* Line 12 */
}
else
{
System.out.println("d");
}
Look closely at line 2, is this an equality check (==) or an assignment (=). The condition at line 2 evaluates to false and also assigns false to bool. bool is now false so the condition at line 6 is not true. The condition at line 10 checks to see if bool is not true ( if !(bool == true) ), it isn't so line 12 is executed.
- hashCode() doesn't have to be overridden if equals() is.
- equals() doesn't have to be overridden if hashCode() is.
- hashCode() can always return the same value, regardless of the object that invoked it.
- equals() can be true even if it's comparing different objects.
(3) and (4) are correct.
(1) and (2) are incorrect because by contract hashCode() and equals() can't be overridden unless both are overridden.
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.
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 */
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".
- start()
- wait()
- notify()
- run()
- terminate()
(1) and (4). Only start() and run() are defined by the Thread class.
(2) and (3) are incorrect because they are methods of the Object class. (5) is incorrect because there's no such method in any thread-related class.
Runnable target = new MyRunnable();
Thread myThread = new Thread(target);
The class correctly implements the Runnable interface with a legal public void run() method.
Option A is incorrect because interfaces are not extended; they are implemented.
Option B is incorrect because even though the class would compile and it has a valid public void run() method, it does not implement the Runnable interface, so the compiler would complain when creating a Thread with an instance of it.
Option D is incorrect because the run() method must be public.
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.
All objects are placed in the garbage collectible heap.
Option A is incorrect because the garbage collector makes no guarantees.
Option B is incorrect because islands of isolated objects can exist.
Option C is incorrect because finalize() has no such mystical powers.
String x = "xyz";
x.toUpperCase(); /* Line 2 */
String y = x.replace('Y', 'y');
y = y + "abc";
System.out.println(y);
Line 2 creates a new String object with the value "XYZ", but this new object is immediately lost because there is no reference to it. Line 3 creates a new String object referenced by y. This new String object has the value "xyz" because there was no "Y" in the String object referred to by x. Line 4 creates a new String object, appends "abc" to the value "xyz", and refers y to the result.
public class Test
{
public static void main(String[] args)
{
final StringBuffer a = new StringBuffer();
final StringBuffer b = new StringBuffer();
new Thread()
{
public void run()
{
System.out.print(a.append("A"));
synchronized(b)
{
System.out.print(b.append("B"));
}
}
}.start();
new Thread()
{
public void run()
{
System.out.print(b.append("C"));
synchronized(a)
{
System.out.print(a.append("D"));
}
}
}.start();
}
}
It gives different output while executing the same compiled code at different times.
C:\>javac Test.java C:\>java Test ABBCAD C:\>java Test ACADCB C:\>java Test ACBCBAD C:\>java Test ABBCAD C:\>java Test ACBCBAD C:\>java Test ACBCBAD C:\>java Test ABBCAD