Discussion
Home ‣ Java Programming ‣ Flow Control See What Others Are Saying!
- Question
What will be the output of the program?
int i = 1, j = 10; do { if(i > j) { break; } j--; } while (++i < 5); System.out.println("i = " + i + " and j = " + j);
Options- A. i = 6 and j = 5
- B. i = 5 and j = 5
- C. i = 6 and j = 4
- D. i = 5 and j = 6
- Correct Answer
- i = 5 and j = 6
ExplanationThis loop is a do-while loop, which always executes the code block within the block at least once, due to the testing condition being at the end of the loop, rather than at the beginning. This particular loop is exited prematurely if i becomes greater than j.The order is, test i against j, if bigger, it breaks from the loop, decrements j by one, and then tests the loop condition, where a pre-incremented by one i is tested for being lower than 5. The test is at the end of the loop, so i can reach the value of 5 before it fails. So it goes, start:
1, 10
2, 9
3, 8
4, 7
5, 6 loop condition fails.
More questions
- 1. Which four options describe the correct default values for array elements of the types indicated?
- int -> 0
- String -> "null"
- Dog -> null
- char -> '\u0000'
- float -> 0.0f
- boolean -> true
Options- A. 1, 2, 3, 4
- B. 1, 3, 4, 5
- C. 2, 4, 5, 6
- D. 3, 4, 5, 6 Discuss
Correct Answer: 1, 3, 4, 5
Explanation:
(1), (3), (4), (5) are the correct statements.(2) is wrong because the default value for a String (and any other object reference) is null, with no quotes.
(6) is wrong because the default value for boolean elements is false.
- 2. Which one of these lists contains only Java programming language keywords?
Options- A. class, if, void, long, Int, continue
- B. goto, instanceof, native, finally, default, throws
- C. try, virtual, throw, final, volatile, transient
- D. strictfp, constant, super, implements, do
- E. byte, break, assert, switch, include Discuss
Correct Answer: goto, instanceof, native, finally, default, throws
Explanation:
All the words in option B are among the 49 Java keywords. Although goto reserved as a keyword in Java, goto is not used and has no function.Option A is wrong because the keyword for the primitive int starts with a lowercase i.
Option C is wrong because "virtual" is a keyword in C++, but not Java.
Option D is wrong because "constant" is not a keyword. Constants in Java are marked static and final.
Option E is wrong because "include" is a keyword in C, but not in Java.
- 3. Which three are legal array declarations?
- int [] myScores [];
- char [] myChars;
- int [6] myScores;
- Dog myDogs [];
- Dog myDogs [7];
Options- A. 1, 2, 4
- B. 2, 4, 5
- C. 2, 3, 4
- D. All are correct. Discuss
Correct Answer: 1, 2, 4
Explanation:
(1), (2), and (4) are legal array declarations. With an array declaration, you can place the brackets to the right or left of the identifier. Option A looks strange, but it's perfectly legal to split the brackets in a multidimensional array, and place them on both sides of the identifier. Although coding this way would only annoy your fellow programmers, for the exam, you need to know it's legal.(3) and (5) are wrong because you can't declare an array with a size. The size is only needed when the array is actually instantiated (and the JVM needs to know how much space to allocate for the array, based on the type of array and the size).
- 4. What will be the output of the program?
class SC2 { public static void main(String [] args) { SC2 s = new SC2(); s.start(); } void start() { int a = 3; int b = 4; System.out.print(" " + 7 + 2 + " "); System.out.print(a + b); System.out.print(" " + a + b + " "); System.out.print(foo() + a + b + " "); System.out.println(a + b + foo()); } String foo() { return "foo"; } }
Options- A. 9 7 7 foo 7 7foo
- B. 72 34 34 foo34 34foo
- C. 9 7 7 foo34 34foo
- D. 72 7 34 foo34 7foo Discuss
Correct Answer: 72 7 34 foo34 7foo
Explanation:
Because all of these expressions use the + operator, there is no precedence to worry about and all of the expressions will be evaluated from left to right. If either operand being evaluated is a String, the + operator will concatenate the two operands; if both operands are numeric, the + operator will add the two operands.- 5. What will be the output of the program?
class Test { public static void main(String [] args) { int x= 0; int y= 0; for (int z = 0; z < 5; z++) { if (( ++x > 2 ) || (++y > 2)) { x++; } } System.out.println(x + " " + y); } }
Options- A. 5 3
- B. 8 2
- C. 8 3
- D. 8 5 Discuss
Correct Answer: 8 2
Explanation:
The first two iterations of the for loop both x and y are incremented. On the third iteration x is incremented, and for the first time becomes greater than 2. The short circuit or operator || keeps y from ever being incremented again and x is incremented twice on each of the last three iterations.- 6. What will be the output of the program?
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.printIn(i); } }
Options- A. 2
- B. 4
- C. 8
- D. 16 Discuss
Correct Answer: 4
Explanation:
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.
- 7. What will be the output of the program?
class PassS { public static void main(String [] args) { PassS p = new PassS(); p.start(); } void start() { String s1 = "slip"; String s2 = fix(s1); System.out.println(s1 + " " + s2); } String fix(String s1) { s1 = s1 + "stream"; System.out.print(s1 + " "); return "stream"; } }
Options- A. slip stream
- B. slipstream stream
- C. stream slip stream
- D. slipstream slip stream Discuss
Correct Answer: slipstream slip stream
Explanation:
When the fix() method is first entered, start()'s s1 and fix()'s s1 reference variables both refer to the same String object (with a value of "slip"). Fix()'s s1 is reassigned to a new object that is created when the concatenation occurs (this second String object has a value of "slipstream"). When the program returns to start(), another String object is created, referred to by s2 and with a value of "stream".- 8. 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 Discuss
Correct Answer: 15 15
Explanation:
Output: 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
- 9. What will be the output of the program?
try { int x = 0; int y = 5 / x; } catch (Exception e) { System.out.println("Exception"); } catch (ArithmeticException ae) { System.out.println(" Arithmetic Exception"); } System.out.println("finished");
Options- A. finished
- B. Exception
- C. Compilation fails.
- D. Arithmetic Exception Discuss
Correct Answer: Compilation fails.
Explanation:
Compilation fails because ArithmeticException has already been caught. ArithmeticException is a subclass of java.lang.Exception, by time the ArithmeticException has been specified it has already been caught by the Exception class.If ArithmeticException appears before Exception, then the file will compile. When catching exceptions the more specific exceptions must be listed before the more general (the subclasses must be caught before the superclasses).
- 10. What will be the output of the program?
public class Test { public static void aMethod() throws Exception { try /* Line 5 */ { throw new Exception(); /* Line 7 */ } finally /* Line 9 */ { System.out.print("finally "); /* Line 11 */ } } public static void main(String args[]) { try { aMethod(); } catch (Exception e) /* Line 20 */ { System.out.print("exception "); } System.out.print("finished"); /* Line 24 */ } }
Options- A. finally
- B. exception finished
- C. finally exception finished
- D. Compilation fails Discuss
Correct Answer: finally exception finished
Explanation:
This is what happens:(1) The execution of the try block (line 5) completes abruptly because of the throw statement (line 7).
(2) The exception cannot be assigned to the parameter of any catch clause of the try statement therefore the finally block is executed (line 9) and "finally" is output (line 11).
(3) The finally block completes normally, and then the try statement completes abruptly because of the throw statement (line 7).
(4) The exception is propagated up the call stack and is caught by the catch in the main method (line 20). This prints "exception".
(5) Lastly program execution continues, because the exception has been caught, and "finished" is output (line 24).
Comments
There are no comments.
Programming
Copyright ©CuriousTab. All rights reserved.
- 1. Which four options describe the correct default values for array elements of the types indicated?