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.
The word "native" is a valid keyword, used to modify a method declaration.
Option A, D and E are not keywords. Option C is wrong because the keyword for subclassing in Java is extends, not 'subclasses'.
public class X
{
public static void main(String [] args)
{
String names [] = new String[5];
for (int x=0; x < args.length; x++)
names[x] = args[x];
System.out.println(names[2]);
}
}
and the command line invocation is
> java X a b
public class CommandArgs
{
public static void main(String [] args)
{
String s1 = args[1];
String s2 = args[2];
String s3 = args[3];
String s4 = args[4];
System.out.print(" args[2] = " + s2);
}
}
and the command-line invocation is
> java CommandArgs 1 2 3 4
An exception is thrown because in the code String s4 = args[4];, the array index (the fifth element) is out of bounds. The exception thrown is the cleverly named ArrayIndexOutOfBoundsException.
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.
public class Test
{
public static void leftshift(int i, int j)
{
i <<= j;
}
public static void main(String args[])
{
int i = 4, j = 2;
leftshift(i, j);
System.out.println(i);
}
}
Java only ever passes arguments to a method by value (i.e. a copy of the variable) and never by reference. Therefore the value of the variable i remains unchanged in the main method.
If you are clever you will spot that 16 is 4 multiplied by 2 twice, (4 * 2 * 2) = 16. If you had 16 left shifted by three bits then 16 * 2 * 2 * 2 = 128. If you had 128 right shifted by 2 bits then 128 / 2 / 2 = 32. Keeping these points in mind, you don't have to go converting to binary to do the left and right bit shifts.
class Test
{
static int s;
public static void main(String [] args)
{
Test p = new Test();
p.start();
System.out.println(s);
}
void start()
{
int x = 7;
twice(x);
System.out.print(x + " ");
}
void twice(int x)
{
x = x*2;
s = x;
}
}
The int x in the twice() method is not the same int x as in the start() method. Start()'s x is not affected by the twice() method. The instance variable s is updated by twice()'s x, which is 14.
- 16*4
- 16>>2
- 16/2^2
- 16>>>2
(2) is correct. 16 >> 2 = 4
(4) is correct. 16 >>> 2 = 4
(1) is wrong. 16 * 4 = 64
(3) is wrong. 16/2 ^ 2 = 10
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).
for(int i = 0; i < 3; i++)
{
switch(i)
{
case 0: break;
case 1: System.out.print("one ");
case 2: System.out.print("two ");
case 3: System.out.print("three ");
}
}
System.out.println("done");
The variable i will have the values 0, 1 and 2.
When i is 0, nothing will be printed because of the break in case 0.
When i is 1, "one two three" will be output because case 1, case 2 and case 3 will be executed (they don't have break statements).
When i is 2, "two three" will be output because case 2 and case 3 will be executed (again no break statements).
Finally, when the for loop finishes "done" will be output.
int I = 0;
label:
if (I < 2) {
System.out.print("I is " + I);
I++;
continue label;
}
The code will not compile because a continue statement can only occur in a looping construct. If this syntax were legal, the combination of the continue and the if statements would create a kludgey kind of loop, but the compiler will force you to write cleaner code than this.
public class Test
{
public static void main (String[] args)
{
String foo = args[1];
String bar = args[2];
String baz = args[3];
System.out.println("baz = " + baz); /* Line 8 */
}
}
And the command line invocation:
> java Test red green blue
When running the program you entered 3 arguments "red", "green" and "blue". When dealing with arrays in java you must remember ALL ARRAYS IN JAVA ARE ZERO BASED therefore args[0] becomes "red", args[1] becomes "green" and args[2] becomes "blue".
When the program entcounters line 8 above at runtime it looks for args[3] which has never been created therefore you get an
ArrayIndexOutOfBoundsException at runtime.
class Foo
{
class Bar{ }
}
class Test
{
public static void main (String [] args)
{
Foo f = new Foo();
/* Line 10: Missing statement ? */
}
}
Option B is correct because the syntax is correct-using both names (the enclosing class and the inner class) in the reference declaration, then using a reference to the enclosing class to invoke new on the inner class.
Option A, C and D all use incorrect syntax. A is incorrect because it doesn't use a reference to the enclosing class, and also because it includes both names in the new.
C is incorrect because it doesn't use the enclosing class name in the reference variable declaration, and because the new syntax is wrong.
D is incorrect because it doesn't use the enclosing class name in the reference variable declaration.
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.
- 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.
class Happy extends Thread
{
final StringBuffer sb1 = new StringBuffer();
final StringBuffer sb2 = new StringBuffer();
public static void main(String args[])
{
final Happy h = new Happy();
new Thread()
{
public void run()
{
synchronized(this)
{
h.sb1.append("A");
h.sb2.append("B");
System.out.println(h.sb1);
System.out.println(h.sb2);
}
}
}.start();
new Thread()
{
public void run()
{
synchronized(this)
{
h.sb1.append("D");
h.sb2.append("C");
System.out.println(h.sb2);
System.out.println(h.sb1);
}
}
}.start();
}
}
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.
public class Test
{
public static void main(String[] args)
{
int x = 0;
assert (x > 0) ? "assertion failed" : "assertion passed" ;
System.out.println("finished");
}
}
Compilation Fails. You can't use the Assert statement in a similar way to the ternary operator. Don't confuse.
What is the value of "d" after this line of code has been executed?
double d = Math.round ( 2.5 + Math.random() );
The Math.random() method returns a number greater than or equal to 0 and less than 1 . Since we can then be sure that the sum of that number and 2.5 will be greater than or equal to 2.5 and less than 3.5, we can be sure that Math.round() will round that number to 3. So Option B is the answer.
public class Test178
{
public static void main(String[] args)
{
String s = "foo";
Object o = (Object)s;
if (s.equals(o))
{
System.out.print("AAA");
}
else
{
System.out.print("BBB");
}
if (o.equals(s))
{
System.out.print("CCC");
}
else
{
System.out.print("DDD");
}
}
}
String x = new String("xyz");
String y = "abc";
x = x + y;
Line 1 creates two, one referred to by x and the lost String "xyz". Line 2 creates one (for a total of three). Line 3 creates one more (for a total of four), the concatenated String referred to by x with a value of "xyzabc".
- If x and y refer to instances of different wrapper classes, then the fragment x.equals(y) will cause a compiler failure.
- If x and y refer to instances of different wrapper classes, then x == y can sometimes be true.
- If x and y are String references and if x.equals(y) is true, then x == y is true.
- If x, y, and z refer to instances of wrapper classes and x.equals(y) is true, and y.equals(z) is true, then z.equals(x) will always be true.
- If x and y are String references and x == y is true, then y.equals(x) will be true.
Statement (4) describes an example of the equals() method behaving transitively. By the way, x, y,and z will all be the same type of wrapper. Statement (5) is true because x and y are referring to the same String object.
Statement (1) is incorrect—the fragment will compile. Statement (2) is incorrect because x == y means that the two reference variables are referring to the same object. Statement (3) will only be true if x and y refer to the same String. It is possible for x and y to refer to two different String objects with the same value.