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- Question
What will be the output of the program?
class PassA { public static void main(String [] args) { PassA p = new PassA(); p.start(); } void start() { long [] a1 = {3,4,5}; long [] a2 = fix(a1); System.out.print(a1[0] + a1[1] + a1[2] + " "); System.out.println(a2[0] + a2[1] + a2[2]); } long [] fix(long [] a3) { a3[1] = 7; return a3; } }
Options- A. 12 15
- B. 15 15
- C. 3 4 5 3 7 5
- D. 3 7 5 3 7 5
- Correct Answer
- 15 15
ExplanationOutput: 15 15The reference variables a1 and a3 refer to the same long array object. When the [1] element is updated in the fix() method, it is updating the array referred to by a1. The reference variable a2 refers to the same array object.
So Output: 3+7+5+" "3+7+5
Output: 15 15 Because Numeric values will be added
More questions
- 1. What will be the output of the program?
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");
Options- A. done
- B. one two done
- C. one two three done
- D. one two three two three done Discuss
Correct Answer: one two three two three done
Explanation:
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.
- 2. Which class does not override the equals() and hashCode() methods, inheriting them directly from class Object?
Options- A. java.lang.String
- B. java.lang.Double
- C. java.lang.StringBuffer
- D. java.lang.Character Discuss
Correct Answer: java.lang.StringBuffer
Explanation:
java.lang.StringBuffer is the only class in the list that uses the default methods provided by class Object.- 3. What will be the output of the program?
public class Delta { static boolean foo(char c) { System.out.print(c); return true; } public static void main( String[] argv ) { int i = 0; for (foo('A'); foo('B') && (i < 2); foo('C')) { i++; foo('D'); } } }
Options- A. ABDCBDCB
- B. ABCDABCD
- C. Compilation fails.
- D. An exception is thrown at runtime. Discuss
Correct Answer: ABDCBDCB
Explanation:
'A' is only printed once at the very start as it is in the initialisation section of the for loop. The loop will only initialise that once.'B' is printed as it is part of the test carried out in order to run the loop.
'D' is printed as it is in the loop.
'C' is printed as it is in the increment section of the loop and will 'increment' only at the end of each loop. Here ends the first loop. Again 'B' is printed as part of the loop test.
'D' is printed as it is in the loop.
'C' is printed as it 'increments' at the end of each loop.
Again 'B' is printed as part of the loop test. At this point the test fails because the other part of the test (i < 2) is no longer true. i has been increased in value by 1 for each loop with the line: i++;
This results in a printout of ABDCBDCB
- 4. What will be the output of the program?
public class Test { public static void main (String [] args) { final Foo f = new Foo(); Thread t = new Thread(new Runnable() { public void run() { f.doStuff(); } }); Thread g = new Thread() { public void run() { f.doStuff(); } }; t.start(); g.start(); } } class Foo { int x = 5; public void doStuff() { if (x < 10) { // nothing to do try { wait(); } catch(InterruptedException ex) { } } else { System.out.println("x is " + x++); if (x >= 10) { notify(); } } } }
Options- A. The code will not compile because of an error on notify(); of class Foo.
- B. The code will not compile because of some other error in class Test.
- C. An exception occurs at runtime.
- D. It prints "x is 5 x is 6". Discuss
Correct Answer: An exception occurs at runtime.
Explanation:
C is correct because the thread does not own the lock of the object it invokes wait() on. If the method were synchronized, the code would run without exception.A, B are incorrect because the code compiles without errors.
D is incorrect because the exception is thrown before there is any output.
- 5. What will be the output of the program?
public class NFE { public static void main(String [] args) { String s = "42"; try { s = s.concat(".5"); /* Line 8 */ double d = Double.parseDouble(s); s = Double.toString(d); int x = (int) Math.ceil(Double.valueOf(s).doubleValue()); System.out.println(x); } catch (NumberFormatException e) { System.out.println("bad number"); } } }
Options- A. 42
- B. 42.5
- C. 43
- D. bad number Discuss
Correct Answer: 43
Explanation:
All of this code is legal, and line 8 creates a new String with a value of "42.5". Lines 9 and 10 convert the String to a double and then back again. Line 11 is fun? Math.ceil()'s argument expression is evaluated first. We invoke the valueOf() method that returns an anonymous Double object (with a value of 42.5). Then the doubleValue() method is called (invoked on the newly created Double object), and returns a double primitive (there and back again), with a value of (you guessed it) 42.5. The ceil() method converts this to 43.0, which is cast to an int and assigned to x.- 6. Which constructs an anonymous inner class instance?
Options- A. Runnable r = new Runnable() { };
- B. Runnable r = new Runnable(public void run() { });
- C. Runnable r = new Runnable { public void run(){}};
- D. System.out.println(new Runnable() {public void run() { }}); Discuss
Correct Answer: System.out.println(new Runnable() {public void run() { }});
Explanation:
D is correct. It defines an anonymous inner class instance, which also means it creates an instance of that new anonymous class at the same time. The anonymous class is an implementer of the Runnable interface, so it must override the run() method of Runnable.A is incorrect because it doesn't override the run() method, so it violates the rules of interface implementation.
B and C use incorrect syntax.
- 7. What is the name of the method used to start a thread execution?
Options- A. init();
- B. start();
- C. run();
- D. resume(); Discuss
Correct Answer: start();
Explanation:
Option B is Correct. The start() method causes this thread to begin execution; the Java Virtual Machine calls the run method of this thread.Option A is wrong. There is no init() method in the Thread class.
Option C is wrong. The run() method of a thread is like the main() method to an application. Starting the thread causes the object's run method to be called in that separately executing thread.
Option D is wrong. The resume() method is deprecated. It resumes a suspended thread.
- 8. What will be the output of the program?
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
Options- A. 0 1 2
- B. 1 2 3
- C. 0 0 0
- D. An exception is thrown at runtime Discuss
Correct Answer: An exception is thrown at runtime
Explanation:
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.- 9. What will be the output of the program?
int i = l, j = -1; switch (i) { case 0, 1: j = 1; /* Line 4 */ case 2: j = 2; default: j = 0; } System.out.println("j = " + j);
Options- A. j = -1
- B. j = 0
- C. j = 1
- D. Compilation fails. Discuss
Correct Answer: Compilation fails.
Explanation:
The case statement takes only a single argument. The case statement on line 4 is given two arguments so the compiler complains.- 10. What will be the output of the program?
public class BoolTest { public static void main(String [] args) { int result = 0; Boolean b1 = new Boolean("TRUE"); Boolean b2 = new Boolean("true"); Boolean b3 = new Boolean("tRuE"); Boolean b4 = new Boolean("false"); if (b1 == b2) /* Line 10 */ result = 1; if (b1.equals(b2) ) /* Line 12 */ result = result + 10; if (b2 == b4) /* Line 14 */ result = result + 100; if (b2.equals(b4) ) /* Line 16 */ result = result + 1000; if (b2.equals(b3) ) /* Line 18 */ result = result + 10000; System.out.println("result = " + result); } }
Options- A. 0
- B. 1
- C. 10
- D. 10010 Discuss
Correct Answer: 10010
Explanation:
Line 10 fails because b1 and b2 are two different objects. Lines 12 and 18 succeed because the Boolean String constructors are case insensitive. Lines 14 and 16 fail because true is not equal to false.
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- 1. What will be the output of the program?