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- Question
What will be the output of the program?
interface Foo141 { int k = 0; /* Line 3 */ } public class Test141 implements Foo141 { public static void main(String args[]) { int i; Test141 test141 = new Test141(); i = test141.k; /* Line 11 */ i = Test141.k; i = Foo141.k; } }
Options- A. Compilation fails.
- B. Compiles and runs ok.
- C. Compiles but throws an Exception at runtime.
- D. Compiles but throws a RuntimeException at runtime.
- Correct Answer
- Compiles and runs ok.
ExplanationThe variable k on line 3 is an interface constant, it is implicitly public, static, and final. Static variables can be referenced in two ways:Via a reference to any instance of the class (line 11)
Via the class name (line 12).
More questions
- 1. Which one is a valid declaration of a boolean?
Options- A. boolean b1 = 0;
- B. boolean b2 = 'false';
- C. boolean b3 = false;
- D. boolean b4 = Boolean.false();
- E. boolean b5 = no; Discuss
Correct Answer: boolean b3 = false;
Explanation:
A boolean can only be assigned the literal true or false.- 2. What will be the output of the program?
String a = "ABCD"; String b = a.toLowerCase(); b.replace('a','d'); b.replace('b','c'); System.out.println(b);
Options- A. abcd
- B. ABCD
- C. dccd
- D. dcba Discuss
Correct Answer: abcd
Explanation:
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');
- 3. What will be the output of the program?
class Q207 { public static void main(String[] args) { int i1 = 5; int i2 = 6; String s1 = "7"; System.out.println(i1 + i2 + s1); /* Line 8 */ } }
Options- A. 18
- B. 117
- C. 567
- D. Compiler error Discuss
Correct Answer: 117
Explanation:
This question is about the + (plus) operator and the overriden + (string cocatanation) operator. The rules that apply when you have a mixed expression of numbers and strings are:If either operand is a String, the + operator concatenates the operands.
If both operands are numeric, the + operator adds the operands.
The expression on line 6 above can be read as "Add the values i1 and i2 together, then take the sum and convert it to a string and concatenate it with the String from the variable s1". In code, the compiler probably interprets the expression on line 8 above as:
System.out.println( new StringBuffer() .append(new Integer(i1 + i2).toString()) .append(s1) .toString() );- 4. In the given program, how many lines of output will be produced?
public class Test { public static void main(String [] args) { int [] [] [] x = new int [3] [] []; int i, j; x[0] = new int[4][]; x[1] = new int[2][]; x[2] = new int[5][]; for (i = 0; i < x.length; i++) { for (j = 0; j < x[i].length; j++) { x[i][j] = new int [i + j + 1]; System.out.println("size = " + x[i][j].length); } } } }
Options- A. 7
- B. 9
- C. 11
- D. 13
- E. Compilation fails Discuss
Correct Answer: 11
Explanation:
The loops use the array sizes (length).It produces 11 lines of output as given below.
D:\Java>javac Test.java D:\Java>java Test size = 1 size = 2 size = 3 size = 4 size = 2 size = 3 size = 3 size = 4 size = 5 size = 6 size = 7
Therefore, 11 is the answer.
- 5. Which of the following statements about the hashcode() method are incorrect?
- The value returned by hashcode() is used in some collection classes to help locate objects.
- The hashcode() method is required to return a positive int value.
- The hashcode() method in the String class is the one inherited from Object.
- Two new empty String objects will produce identical hashcodes.
Options- A. 1 and 2
- B. 2 and 3
- C. 3 and 4
- D. 1 and 4 Discuss
Correct Answer: 2 and 3
Explanation:
(2) is an incorrect statement because there is no such requirement.(3) is an incorrect statement and therefore a correct answer because the hashcode for a string is computed from the characters in the string.
- 6. What will be the output of the program?
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
Options- A. names
- B. null
- C. Compilation fails
- D. An exception is thrown at runtime Discuss
Correct Answer: null
Explanation:
The names array is initialized with five null elements. Then elements 0 and 1 are assigned the String values "a" and "b" respectively (the command-line arguments passed to main). Elements of names array 2, 3, and 4 remain unassigned, so they have a value of null.- 7. What will be the output of the program?
class MyThread extends Thread { public static void main(String [] args) { MyThread t = new MyThread(); t.start(); System.out.print("one. "); t.start(); System.out.print("two. "); } public void run() { System.out.print("Thread "); } }
Options- A. Compilation fails
- B. An exception occurs at runtime.
- C. It prints "Thread one. Thread two."
- D. The output cannot be determined. Discuss
Correct Answer: An exception occurs at runtime.
Explanation:
When the start() method is attempted a second time on a single Thread object, the method will throw an IllegalThreadStateException (you will not need to know this exception name for the exam). Even if the thread has finished running, it is still illegal to call start() again.- 8. What will be the output of the program?
public class ExamQuestion7 { static int j; static void methodA(int i) { boolean b; do { b = i<10 | methodB(4); /* Line 9 */ b = i<10 || methodB(8); /* Line 10 */ }while (!b); } static boolean methodB(int i) { j += i; return true; } public static void main(String[] args) { methodA(0); System.out.println( "j = " + j ); } }
Options- A. j = 0
- B. j = 4
- C. j = 8
- D. The code will run with no output Discuss
Correct Answer: j = 4
Explanation:
The lines to watch here are lines 9 & 10. Line 9 features the non-shortcut version of the OR operator so both of its operands will be evaluated and therefore methodB(4) is executed.However line 10 has the shortcut version of the OR operator and if the 1st of its operands evaluates to true (which in this case is true), then the 2nd operand isn't evaluated, so methodB(8) never gets called.
The loop is only executed once, b is initialized to false and is assigned true on line 9. Thus j = 4.
- 9. The following block of code creates a Thread using a Runnable target:
Runnable target = new MyRunnable(); Thread myThread = new Thread(target);
Which of the following classes can be used to create the target, so that the preceding code compiles correctly?
Options- A. public class MyRunnable extends Runnable{public void run(){}}
- B. public class MyRunnable extends Object{public void run(){}}
- C. public class MyRunnable implements Runnable{public void run(){}}
- D. public class MyRunnable implements Runnable{void run(){}} Discuss
Correct Answer: public class MyRunnable implements Runnable{public void run(){}}
Explanation:
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.
- 10. What will be the output of the program?
public class WaitTest { public static void main(String [] args) { System.out.print("1 "); synchronized(args) { System.out.print("2 "); try { args.wait(); /* Line 11 */ } catch(InterruptedException e){ } } System.out.print("3 "); } }
Options- A. It fails to compile because the IllegalMonitorStateException of wait() is not dealt with in line 11.
- B. 1 2 3
- C. 1 3
- D. 1 2 Discuss
Correct Answer: 1 2
Explanation:
1 and 2 will be printed, but there will be no return from the wait call because no other thread will notify the main thread, so 3 will never be printed. The program is essentially frozen at line 11.A is incorrect; IllegalMonitorStateException is an unchecked exception so it doesn't have to be dealt with explicitly.
B and C are incorrect; 3 will never be printed, since this program will never terminate because it will wait forever.
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