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.
All the words in option B are among the 49 Java keywords. Although goto reserved as a keyword in Java, goto is not used and has no function.
Option A is wrong because the keyword for the primitive int starts with a lowercase i.
Option C is wrong because "virtual" is a keyword in C++, but not Java.
Option D is wrong because "constant" is not a keyword. Constants in Java are marked static and final.
Option E is wrong because "include" is a keyword in C, but not in Java.
Option A is correct. A public access modifier is acceptable. The method prototypes in an interface are all abstract by virtue of their declaration, and should not be declared abstract.
Option B is wrong. The final modifier means that this method cannot be constructed in a subclass. A final method cannot be abstract.
Option C is wrong. static is concerned with the class and not an instance.
Option D is wrong. protected is not permitted when declaring a method of an interface. See information below.
Member declarations in an interface disallow the use of some declaration modifiers; you cannot use transient, volatile, or synchronized in a member declaration in an interface. Also, you may not use the private and protected specifiers when declaring members of an interface.
public class CommandArgsTwo
{
public static void main(String [] argh)
{
int x;
x = argh.length;
for (int y = 1; y <= x; y++)
{
System.out.print(" " + argh[y]);
}
}
}
and the command-line invocation is
> java CommandArgsTwo 1 2 3
An exception is thrown because at some point in (System.out.print(" " + argh[y]);), the value of x will be equal to y, resulting in an attempt to access an index out of bounds for the array. Remember that you can access only as far as length - 1, so loop logical tests should use x < someArray.length as opposed to x < = someArray.length.
- 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.
- 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 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 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.
import java.util.*;
public class NewTreeSet2 extends NewTreeSet
{
public static void main(String [] args)
{
NewTreeSet2 t = new NewTreeSet2();
t.count();
}
}
protected class NewTreeSet
{
void count()
{
for (int x = 0; x < 7; x++,x++ )
{
System.out.print(" " + x);
}
}
}
Nonnested classes cannot be marked protected (or final for that matter), so the compiler will fail at protected class NewTreeSet.
/* Missing statements ? */
public class NewTreeSet extends java.util.TreeSet
{
public static void main(String [] args)
{
java.util.TreeSet t = new java.util.TreeSet();
t.clear();
}
public void clear()
{
TreeMap m = new TreeMap();
m.clear();
}
}
- No statement is required
- import java.util.*;
- import.java.util.Tree*;
- import java.util.TreeSet;
- import java.util.TreeMap;
(2) and (5). TreeMap is the only class that must be imported. TreeSet does not need an import statement because it is described with a fully qualified name.
(1) is incorrect because TreeMap must be imported. (3) is incorrect syntax for an import statement. (4) is incorrect because it will not import TreeMap, which is required.
- 3/2
- 3<2
- 3*4
- 3<<2
(1) is wrong. 3/2 = 1 (integer arithmetic).
(2) is wrong. 3 < 2 = false.
(3) is correct. 3 * 4 = 12.
(4) is correct. 3 <<2= 12. In binary 3 is 11, now shift the bits two places to the left and we get 1100 which is 12 in binary (3*2*2).
int x = l, y = 6;
while (y--)
{
x++;
}
System.out.println("x = " + x +" y = " + y);
Compilation fails because the while loop demands a boolean argument for it's looping condition, but in the code, it's given an int argument.
while(true) { //insert code here }
public class Switch2
{
final static short x = 2;
public static int y = 0;
public static void main(String [] args)
{
for (int z=0; z < 3; z++)
{
switch (z)
{
case x: System.out.print("0 ");
case x-1: System.out.print("1 ");
case x-2: System.out.print("2 ");
}
}
}
}
The case expressions are all legal because x is marked final, which means the expressions can be evaluated at compile time. In the first iteration of the for loop case x-2 matches, so 2 is printed. In the second iteration, x-1 is matched so 1 and 2 are printed (remember, once a match is found all remaining statements are executed until a break statement is encountered). In the third iteration, x is matched. So 0 1 and 2 are printed.
public class Switch2
{
final static short x = 2;
public static int y = 0;
public static void main(String [] args)
{
for (int z=0; z < 4; z++)
{
switch (z)
{
case x: System.out.print("0 ");
default: System.out.print("def ");
case x-1: System.out.print("1 ");
break;
case x-2: System.out.print("2 ");
}
}
}
}
When z == 0 , case x-2 is matched. When z == 1, case x-1 is matched and then the break occurs. When z == 2, case x, then default, then x-1 are all matched. When z == 3, default, then x-1 are matched. The rules for default are that it will fall through from above like any other case (for instance when z == 2), and that it will match when no other cases match (for instance when z==3).
public class Foo
{
public static void main(String[] args)
{
try
{
return;
}
finally
{
System.out.println( "Finally" );
}
}
}
- An exception arising in the finally block itself.
- The death of the thread.
- The use of System.exit()
- Turning off the power to the CPU.
I suppose the last three could be classified as VM shutdown.
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.
- The Iterator interface declares only three methods: hasNext, next and remove.
- The ListIterator interface extends both the List and Iterator interfaces.
- The ListIterator interface provides forward and backward iteration capabilities.
- The ListIterator interface provides the ability to modify the List during iteration.
- The ListIterator interface provides the ability to determine its position in the List.
The ListIterator interface extends the Iterator interface and declares additional methods to provide forward and backward iteration capabilities, List modification capabilities, and the ability to determine the position of the iterator in the List.
class HappyGarbage01
{
public static void main(String args[])
{
HappyGarbage01 h = new HappyGarbage01();
h.methodA(); /* Line 6 */
}
Object methodA()
{
Object obj1 = new Object();
Object [] obj2 = new Object[1];
obj2[0] = obj1;
obj1 = null;
return obj2[0];
}
}
Option D is correct. Garbage collection takes place after the method has returned its reference to the object. The method returns to line 6, there is no reference to store the return value. so garbage collection takes place after line 6.
Option A is wrong. Because the reference to obj1 is stored in obj2[0]. The Object obj1 still exists on the heap and can be accessed by an active thread through the reference stored in obj2[0].
Option B is wrong. Because it is only one of the references to the object obj1, the other reference is maintained in obj2[0].
Option C is wrong. The garbage collector will not be called here because a reference to the object is being maintained and returned in obj2[0].
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.
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.
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.