Discussion
Home ‣ Java Programming ‣ Garbage Collections See What Others Are Saying!
- Question
What allows the programmer to destroy an object x?
Options- A. x.delete()
- B. x.finalize()
- C. Runtime.getRuntime().gc()
- D. Only the garbage collection system can destroy an object.
- Correct Answer
- Only the garbage collection system can destroy an object.
ExplanationOption 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:
More questions
- 1. Which two statements are equivalent?
- 16*4
- 16>>2
- 16/2^2
- 16>>>2
Options- A. 1 and 2
- B. 2 and 4
- C. 3 and 4
- D. 1 and 3 Discuss
Correct Answer: 2 and 4
Explanation:
(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
- 2. What two statements are true about properly overridden hashCode() and equals() methods?
- 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.
Options- A. 1 and 2
- B. 2 and 3
- C. 3 and 4
- D. 1 and 3 Discuss
Correct Answer: 3 and 4
Explanation:
(3) and (4) are correct.(1) and (2) are incorrect because by contract hashCode() and equals() can't be overridden unless both are overridden.
- 3. Which is a valid keyword in java?
Options- A. interface
- B. string
- C. Float
- D. unsigned Discuss
Correct Answer: interface
Explanation:
interface is a valid keyword.Option B is wrong because although "String" is a class type in Java, "string" is not a keyword.
Option C is wrong because "Float" is a class type. The keyword for the Java primitive is float.
Option D is wrong because "unsigned" is a keyword in C/C++ but not in Java.
- 4. What will be the output of the program?
int i = 1, j = 10; do { if(i++ > --j) /* Line 4 */ { continue; } } while (i < 5); System.out.println("i = " + i + "and j = " + j); /* Line 9 */
Options- A. i = 6 and j = 5
- B. i = 5 and j = 5
- C. i = 6 and j = 6
- D. i = 5 and j = 6 Discuss
Correct Answer: i = 5 and j = 6
Explanation:
This question is not testing your knowledge of the continue statement. It is testing your knowledge of the order of evaluation of operands. Basically the prefix and postfix unary operators have a higher order of evaluation than the relational operators. So on line 4 the variable i is incremented and the variable j is decremented before the greater than comparison is made. As the loop executes the comparison on line 4 will be:if(i > j)
if(2 > 9)
if(3 > 8)
if(4 > 7)
if(5 > 6) at this point i is not less than 5, therefore the loop terminates and line 9 outputs the values of i and j as 5 and 6 respectively.
The continue statement never gets to execute because i never reaches a value that is greater than j.
- 5. What will be the output of the program?
int i = 0; while(1) { if(i == 4) { break; } ++i; } System.out.println("i = " + i);
Options- A. i = 0
- B. i = 3
- C. i = 4
- D. Compilation fails. Discuss
Correct Answer: Compilation fails.
Explanation:
Compilation fails because the argument of the while loop, the condition, must be of primitive type boolean. In Java, 1 does not represent the true state of a boolean, rather it is seen as an integer.- 6. What will be the output of the program?
public class Test { public int aMethod() { static int i = 0; i++; return i; } public static void main(String args[]) { Test test = new Test(); test.aMethod(); int j = test.aMethod(); System.out.println(j); } }
Options- A. 0
- B. 1
- C. 2
- D. Compilation fails. Discuss
Correct Answer: Compilation fails.
Explanation:
Compilation failed because static was an illegal start of expression - method variables do not have a modifier (they are always considered local).- 7. What will be the output of the program?
int x = l, y = 6; while (y--) { x++; } System.out.println("x = " + x +" y = " + y);
Options- A. x = 6 y = 0
- B. x = 7 y = 0
- C. x = 6 y = -1
- D. Compilation fails. Discuss
Correct Answer: Compilation fails.
Explanation:
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 }
- 8. What will be the output of the program?
class MyThread extends Thread { MyThread() {} MyThread(Runnable r) {super(r); } public void run() { System.out.print("Inside Thread "); } } class MyRunnable implements Runnable { public void run() { System.out.print(" Inside Runnable"); } } class Test { public static void main(String[] args) { new MyThread().start(); new MyThread(new MyRunnable()).start(); } }
Options- A. Prints "Inside Thread Inside Thread"
- B. Prints "Inside Thread Inside Runnable"
- C. Does not compile
- D. Throws exception at runtime Discuss
Correct Answer: Prints "Inside Thread Inside Thread"
Explanation:
If a Runnable object is passed to the Thread constructor, then the run method of the Thread class will invoke the run method of the Runnable object.In this case, however, the run method in the Thread class is overridden by the run method in MyThread class. Therefore the run() method in MyRunnable is never invoked.
Both times, the run() method in MyThread is invoked instead.
- 9. Which statement is true?
Options- A. Assertions can be enabled or disabled on a class-by-class basis.
- B. Conditional compilation is used to allow tested classes to run at full speed.
- C. Assertions are appropriate for checking the validity of arguments in a method.
- D. The programmer can choose to execute a return statement or to throw an exception if an assertion fails. Discuss
Correct Answer: Assertions can be enabled or disabled on a class-by-class basis.
Explanation:
Option A is correct. The assertion status can be set for a named top-level class and any nested classes contained therein. This setting takes precedence over the class loader's default assertion status, and over any applicable per-package default. If the named class is not a top-level class, the change of status will have no effect on the actual assertion status of any class.Option B is wrong. Is there such a thing as conditional compilation in Java?
Option C is wrong. For private methods - yes. But do not use assertions to check the parameters of a public method. An assert is inappropriate in public methods because the method guarantees that it will always enforce the argument checks. A public method must check its arguments whether or not assertions are enabled. Further, the assert construct does not throw an exception of the specified type. It can throw only an AssertionError.
Option D is wrong. Because you're never supposed to handle an assertion failure. That means don't catch it with a catch clause and attempt to recover.
- 10. What will be the output of the program?
class Two { byte x; } class PassO { public static void main(String [] args) { PassO p = new PassO(); p.start(); } void start() { Two t = new Two(); System.out.print(t.x + " "); Two t2 = fix(t); System.out.println(t.x + " " + t2.x); } Two fix(Two tt) { tt.x = 42; return tt; } }
Options- A. null null 42
- B. 0 0 42
- C. 0 42 42
- D. 0 0 0 Discuss
Correct Answer: 0 42 42
Explanation:
In the fix() method, the reference variable tt refers to the same object (class Two) as the t reference variable. Updating tt.x in the fix() method updates t.x (they are one in the same object). Remember also that the instance variable x in the Two class is initialized to 0.
Comments
There are no comments.
Programming
Copyright ©CuriousTab. All rights reserved.