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- Question
Which three guarantee that a thread will leave the running state?
- yield()
- wait()
- notify()
- notifyAll()
- sleep(1000)
- aLiveThread.join()
- Thread.killThread()
Options- A. 1, 2 and 4
- B. 2, 5 and 6
- C. 3, 4 and 7
- D. 4, 5 and 7
- Correct Answer
- 2, 5 and 6
Explanation(2) is correct because wait() always causes the current thread to go into the object's wait pool.(5) is correct because sleep() will always pause the currently running thread for at least the duration specified in the sleep argument (unless an interrupted exception is thrown).
(6) is correct because, assuming that the thread you're calling join() on is alive, the thread calling join() will immediately block until the thread you're calling join() on is no longer alive.
(1) is wrong, but tempting. The yield() method is not guaranteed to cause a thread to leave the running state, although if there are runnable threads of the same priority as the currently running thread, then the current thread will probably leave the running state.
(3) and (4) are incorrect because they don't cause the thread invoking them to leave the running state.
(7) is wrong because there's no such method.
More questions
- 1. Which of the following line of code is suitable to start a thread?
class X implements Runnable { public static void main(String args[]) { /* Missing code? */ } public void run() {} }
Options- A. Thread t = new Thread(X);
- B. Thread t = new Thread(X); t.start();
- C. X run = new X(); Thread t = new Thread(run); t.start();
- D. Thread t = new Thread(); x.run(); Discuss
Correct Answer: X run = new X(); Thread t = new Thread(run); t.start();
Explanation:
Option C is suitable to start a thread.- 2. What will be the output of the program?
try { Float f1 = new Float("3.0"); int x = f1.intValue(); byte b = f1.byteValue(); double d = f1.doubleValue(); System.out.println(x + b + d); } catch (NumberFormatException e) /* Line 9 */ { System.out.println("bad number"); /* Line 11 */ }
Options- A. 9.0
- B. bad number
- C. Compilation fails on line 9.
- D. Compilation fails on line 11. Discuss
Correct Answer: 9.0
Explanation:
The xxxValue() methods convert any numeric wrapper object's value to any primitive type. When narrowing is necessary, significant bits are dropped and the results are difficult to calculate.- 3. What will be the output of the program?
String a = "newspaper"; a = a.substring(5,7); char b = a.charAt(1); a = a + b; System.out.println(a);
Options- A. apa
- B. app
- C. apea
- D. apep Discuss
Correct Answer: app
Explanation:
Both substring() and charAt() methods are indexed with a zero-base, and substring() returns a String of length arg2 - arg1.- 4. Which statement is true?
Options- A. The notifyAll() method must be called from a synchronized context.
- B. To call wait(), an object must own the lock on the thread.
- C. The notify() method is defined in class java.lang.Thread.
- D. The notify() method causes a thread to immediately release its locks. Discuss
Correct Answer: The notifyAll() method must be called from a synchronized context.
Explanation:
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.
- 5. Which statement is true?
class Test1 { public int value; public int hashCode() { return 42; } } class Test2 { public int value; public int hashcode() { return (int)(value^5); } }
Options- A. class Test1 will not compile.
- B. The Test1 hashCode() method is more efficient than the Test2 hashCode() method.
- C. The Test1 hashCode() method is less efficient than the Test2 hashCode() method.
- D. class Test2 will not compile. Discuss
Correct Answer: The Test1 hashCode() method is less efficient than the Test2 hashCode() method.
Explanation:
The so-called "hashing algorithm" implemented by class Test1 will always return the same value, 42, which is legal but which will place all of the hash table entries into a single bucket, the most inefficient setup possible.Option A and D are incorrect because these classes are legal.
Option B is incorrect based on the logic described above.
- 6. Assuming that the equals() and hashCode() methods are properly implemented, if the output is "x = 1111", which of the following statements will always be true?
x = 0; if (x1.hashCode() != x2.hashCode() ) x = x + 1; if (x3.equals(x4) ) x = x + 10; if (!x5.equals(x6) ) x = x + 100; if (x7.hashCode() == x8.hashCode() ) x = x + 1000; System.out.println("x = " + x);
Options- A. x2.equals(x1)
- B. x3.hashCode() == x4.hashCode()
- C. x5.hashCode() != x6.hashCode()
- D. x8.equals(x7) Discuss
Correct Answer: x3.hashCode() == x4.hashCode()
Explanation:
By contract, if two objects are equivalent according to the equals() method, then the hashCode() method must evaluate them to be ==.Option A is incorrect because if the hashCode() values are not equal, the two objects must not be equal.
Option C is incorrect because if equals() is not true there is no guarantee of any result from hashCode().
Option D is incorrect because hashCode() will often return == even if the two objects do not evaluate to equals() being true.
- 7. The static method Thread.currentThread() returns a reference to the currently executing Thread object. What is the result of this code?
class Test { public static void main(String [] args) { printAll(args); } public static void printAll(String[] lines) { for(int i = 0; i < lines.length; i++) { System.out.println(lines[i]); Thread.currentThread().sleep(1000); } } }
Options- A. Each String in the array lines will output, with a 1-second pause.
- B. Each String in the array lines will output, with no pause in between because this method is not executed in a Thread.
- C. Each String in the array lines will output, and there is no guarantee there will be a pause because currentThread() may not retrieve this thread.
- D. This code will not compile. Discuss
Correct Answer: This code will not compile.
Explanation:
D. The sleep() method must be enclosed in a try/catch block, or the method printAll() must declare it throws the InterruptedException.A is incorrect, but it would be correct if the InterruptedException was dealt with.
B is incorrect, but it would still be incorrect if the InterruptedException was dealt with because all Java code, including the main() method, runs in threads.
C is incorrect. The sleep() method is static, so even if it is called on an instance, it still always affects the currently executing thread.
- 8. Which answer most closely indicates the behavior of the program?
public class MyProgram { public static void throwit() { throw new RuntimeException(); } public static void main(String args[]) { try { System.out.println("Hello world "); throwit(); System.out.println("Done with try block "); } finally { System.out.println("Finally executing "); } } }
Options- A. The program will not compile.
- B. The program will print Hello world, then will print that a RuntimeException has occurred, then will print Done with try block, and then will print Finally executing.
- C. The program will print Hello world, then will print that a RuntimeException has occurred, and then will print Finally executing.
- D. The program will print Hello world, then will print Finally executing, then will print that a RuntimeException has occurred. Discuss
Correct Answer: The program will print Hello world, then will print Finally executing, then will print that a RuntimeException has occurred.
Explanation:
Once the program throws a RuntimeException (in the throwit() method) that is not caught, the finally block will be executed and the program will be terminated. If a method does not handle an exception, the finally block is executed before the exception is propagated.- 9. Which two code fragments will compile?
- interface Base2 implements Base {}
- abstract class Class2 extends Base
{ public boolean m1(){ return true; }} - abstract class Class2 implements Base {}
- abstract class Class2 implements Base
{ public boolean m1(){ return (7 > 4); }} - abstract class Class2 implements Base
{ protected boolean m1(){ return (5 > 7) }}
interface Base { boolean m1 (); byte m2(short s); }
Options- A. 1 and 2
- B. 2 and 3
- C. 3 and 4
- D. 1 and 5 Discuss
Correct Answer: 3 and 4
Explanation:
(3) is correct because an abstract class doesn't have to implement any or all of its interface's methods. (4) is correct because the method is correctly implemented ((7 > 4) is a boolean).(1) is incorrect because interfaces don't implement anything. (2) is incorrect because classes don't extend interfaces. (5) is incorrect because interface methods are implicitly public, so the methods being implemented must be public.
- 10. Which statement, if placed in a class other than MyOuter or MyInner, instantiates an instance of the nested class?
public class MyOuter { public static class MyInner { public static void foo() { } } }
Options- A. MyOuter.MyInner m = new MyOuter.MyInner();
- B. MyOuter.MyInner mi = new MyInner();
- C. MyOuter m = new MyOuter();
MyOuter.MyInner mi = m.new MyOuter.MyInner();
- D. MyInner mi = new MyOuter.MyInner(); Discuss
Correct Answer: MyOuter.MyInner m = new MyOuter.MyInner();
Explanation:
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.
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- 1. Which of the following line of code is suitable to start a thread?