Online Java Programming Test - Java Programming Test 3
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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.
- float[ ] f = new float(3);
- float f2[ ] = new float[ ];
- float[ ]f1 = new float[3];
- float f3[ ] = new float[3];
- float f5[ ] = {1.0f, 2.0f, 2.0f};
(1) causes two compiler errors ( '[' expected and illegal start of expression) because the wrong type of bracket is used, ( ) instead of [ ]. The following is the correct syntax: float[ ] f = new float[3];
(2) causes a compiler error ( '{' expected ) because the array constructor does not specify the number of elements in the array. The following is the correct syntax: float f2[ ] = new float[3];
(3), (4), and (5) compile without error.
public class Test { }
Option A and B are wrong because they use the default access modifier and the access modifier for the class is public (remember, the default constructor has the same access modifier as the class).
Option D is wrong. The void makes the compiler think that this is a method specification - in fact if it were a method specification the compiler would spit it out.
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.
public class Test
{
public static void main(String args[])
{
class Foo
{
public int i = 3;
}
Object o = (Object)new Foo();
Foo foo = (Foo)o;
System.out.println("i = " + foo.i);
}
}
public class A
{
void A() /* Line 3 */
{
System.out.println("Class A");
}
public static void main(String[] args)
{
new A();
}
}
Option D is correct. The specification at line 3 is for a method and not a constructor and this method is never called therefore there is no output. The constructor that is called is the default constructor.
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);
}
}
The first two iterations of the for loop both x and y are incremented. On the third iteration x is incremented, and for the first time becomes greater than 2. The short circuit or operator || keeps y from ever being incremented again and x is incremented twice on each of the last three iterations.
import java.awt.Button;
class CompareReference
{
public static void main(String [] args)
{
float f = 42.0f;
float [] f1 = new float[2];
float [] f2 = new float[2];
float [] f3 = f1;
long x = 42;
f1[0] = 42.0f;
}
}
- f1 == f2
- f1 == f3
- f2 == f1[1]
- x == f1[0]
- f == f1[0]
(2) is correct because the reference variables f1 and f3 refer to the same array object.
(4) is correct because it is legal to compare integer and floating-point types.
(5) is correct because it is legal to compare a variable with an array element.
(3) is incorrect because f2 is an array object and f1[1] is an array element.
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.
System.out.print("Start ");
try
{
System.out.print("Hello world");
throw new FileNotFoundException();
}
System.out.print(" Catch Here "); /* Line 7 */
catch(EOFException e)
{
System.out.print("End of file exception");
}
catch(FileNotFoundException e)
{
System.out.print("File not found");
}
Line 7 will cause a compiler error. The only legal statements after try blocks are either catch or finally statements.
Option B, C, and D are incorrect based on the program logic described above. If line 7 was removed, the code would compile and the correct answer would be Option B.
java.lang.StringBuffer is the only class in the list that uses the default methods provided by class Object.
Option B is correct. A set is a collection that contains no duplicate elements. The iterator returns the elements in no particular order (unless this set is an instance of some class that provides a guarantee). A map cannot contain duplicate keys but it may contain duplicate values. List and Collection allow duplicate elements.
Option A is wrong. A map is an object that maps keys to values. A map cannot contain duplicate keys; each key can map to at most one value. The Map interface provides three collection views, which allow a map's contents to be viewed as a set of keys, collection of values, or set of key-value mappings. The order of a map is defined as the order in which the iterators on the map's collection views return their elements. Some map implementations, like the TreeMap class, make specific guarantees as to their order (ascending key order); others, like the HashMap class, do not (does not guarantee that the order will remain constant over time).
Option C is wrong. A list is an ordered collection (also known as a sequence). The user of this interface has precise control over where in the list each element is inserted. The user can access elements by their integer index (position in the list), and search for elements in the list. Unlike sets, lists typically allow duplicate elements.
Option D is wrong. A collection is also known as a sequence. The user of this interface has precise control over where in the list each element is inserted. The user can access elements by their integer index (position in the list), and search for elements in the list. Unlike sets, lists typically allow duplicate elements.
/* 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.
public class MyOuter
{
public static class MyInner
{
public static void foo() { }
}
}
MyInner is a static nested class, so it must be instantiated using the fully-scoped name of MyOuter.MyInner.
Option B is incorrect because it doesn't use the enclosing name in the new.
Option C is incorrect because it uses incorrect syntax. When you instantiate a nested class by invoking new on an instance of the enclosing class, you do not use the enclosing name. The difference between Option A and C is that Option C is calling new on an instance of the enclosing class rather than just new by itself.
Option D is incorrect because it doesn't use the enclosing class name in the variable declaration.
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.
- When an interrupted exception occurs.
- When a thread of higher priority is ready (becomes runnable).
- When the thread creates a new thread.
- When the stop() method is called.
- Thread(Runnable r, String name)
- Thread()
- Thread(int priority)
- Thread(Runnable r, ThreadGroup g)
- Thread(Runnable r, int priority)
(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.
- 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.
Option A is correct because the notifyAll() method (along with wait() and notify()) must always be called from within a synchronized context.
Option B is incorrect because to call wait(), the thread must own the lock on the object that wait() is being invoked on, not the other way around.
Option C is wrong because notify() is defined in java.lang.Object.
Option D is wrong because notify() will not cause a thread to release its locks. The thread can only release its locks by exiting the synchronized code.
public Object m()
{
Object o = new Float(3.14F);
Object [] oa = new Object[l];
oa[0] = o; /* Line 5 */
o = null; /* Line 6 */
oa[0] = null; /* Line 7 */
return o; /* Line 8 */
}
Option A is wrong. This simply copies the object reference into the array.
Option B is wrong. The reference o is set to null, but, oa[0] still maintains the reference to the Float object.
Option C is correct. The thread of execution will then not have access to the object.
String a = "ABCD";
String b = a.toLowerCase();
b.replace('a','d');
b.replace('b','c');
System.out.println(b);
String objects are immutable, they cannot be changed, in this case we are talking about the replace method which returns a new String object resulting from replacing all occurrences of oldChar in this string with newChar.
b.replace(char oldChar, char newChar);
But since this is only a temporary String it must either be put to use straight away i.e.
System.out.println(b.replace('a','d'));
Or a new variable must be assigned its value i.e.
String c = b.replace('a','d');