Discussion
Home ‣ Java Programming ‣ Flow Control Comments
- 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.
Flow Control problems
Search Results
- 1. What will be the output of the program?
class A { final public int GetResult(int a, int b) { return 0; } } class B extends A { public int GetResult(int a, int b) {return 1; } } public class Test { public static void main(String args[]) { B b = new B(); System.out.println("x = " + b.GetResult(0, 1)); } }
Options- A. x = 0
- B. x = 1
- C. Compilation fails.
- D. An exception is thrown at runtime. Discuss
Correct Answer: Compilation fails.
Explanation:
The code doesn't compile because the method GetResult() in class A is final and so cannot be overridden.- 2. What will be the output of the program?
class Base { Base() { System.out.print("Base"); } } public class Alpha extends Base { public static void main(String[] args) { new Alpha(); /* Line 12 */ new Base(); /* Line 13 */ } }
Options- A. Base
- B. BaseBase
- C. Compilation fails
- D. The code runs with no output Discuss
Correct Answer: BaseBase
Explanation:
Option B is correct. It would be correct if the code had compiled, and the subclass Alpha had been saved in its own file. In this case Java supplies an implicit call from the sub-class constructor to the no-args constructor of the super-class therefore line 12 causes Base to be output. Line 13 also causes Base to be output.Option A is wrong. It would be correct if either the main class or the subclass had not been instantiated.
Option C is wrong. The code compiles.
Option D is wrong. There is output.
- 3. What will be the output of the program?
class Super { public int i = 0; public Super(String text) /* Line 4 */ { i = 1; } } class Sub extends Super { public Sub(String text) { i = 2; } public static void main(String args[]) { Sub sub = new Sub("Hello"); System.out.println(sub.i); } }
Options- A. 0
- B. 1
- C. 2
- D. Compilation fails. Discuss
Correct Answer: Compilation fails.
Explanation:
A default no-args constructor is not created because there is a constructor supplied that has an argument, line 4. Therefore the sub-class constructor must explicitly make a call to the super class constructor:- 4. What will be the output of the program?
class Super { public Integer getLength() { return new Integer(4); } } public class Sub extends Super { public Long getLength() { return new Long(5); } public static void main(String[] args) { Super sooper = new Super(); Sub sub = new Sub(); System.out.println( sooper.getLength().toString() + "," + sub.getLength().toString() ); } }
Options- A. 4, 4
- B. 4, 5
- C. 5, 4
- D. Compilation fails. Discuss
Correct Answer: Compilation fails.
Explanation:
Option D is correct, compilation fails - The return type of getLength( ) in the super class is an object of reference type Integer and the return type in the sub class is an object of reference type Long. In other words, it is not an override because of the change in the return type and it is also not an overload because the argument list has not changed.- 5. 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).- 6. What will be the output of the program?
boolean bool = true; if(bool = false) /* Line 2 */ { System.out.println("a"); } else if(bool) /* Line 6 */ { System.out.println("b"); } else if(!bool) /* Line 10 */ { System.out.println("c"); /* Line 12 */ } else { System.out.println("d"); }
Options- A. a
- B. b
- C. c
- D. d Discuss
Correct Answer: c
Explanation:
Look closely at line 2, is this an equality check (==) or an assignment (=). The condition at line 2 evaluates to false and also assigns false to bool. bool is now false so the condition at line 6 is not true. The condition at line 10 checks to see if bool is not true ( if !(bool == true) ), it isn't so line 12 is executed.- 7. What will be the output of the program?
public class If2 { static boolean b1, b2; public static void main(String [] args) { int x = 0; if ( !b1 ) /* Line 7 */ { if ( !b2 ) /* Line 9 */ { b1 = true; x++; if ( 5 > 6 ) { x++; } if ( !b1 ) x = x + 10; else if ( b2 = true ) /* Line 19 */ x = x + 100; else if ( b1 | b2 ) /* Line 21 */ x = x + 1000; } } System.out.println(x); } }
Options- A. 0
- B. 1
- C. 101
- D. 111 Discuss
Correct Answer: 101
Explanation:
As instance variables, b1 and b2 are initialized to false. The if tests on lines 7 and 9 are successful so b1 is set to true and x is incremented. The next if test to succeed is on line 19 (note that the code is not testing to see if b2 is true, it is setting b2 to be true). Since line 19 was successful, subsequent else-if's (line 21) will be skipped.- 8. What will be the output of the program?
int I = 0; label: if (I < 2) { System.out.print("I is " + I); I++; continue label; }
Options- A. I is 0
- B. I is 0 I is 1
- C. Compilation fails.
- D. None of the above Discuss
Correct Answer: Compilation fails.
Explanation:
The code will not compile because a continue statement can only occur in a looping construct. If this syntax were legal, the combination of the continue and the if statements would create a kludgey kind of loop, but the compiler will force you to write cleaner code than this.- 9. What will be the output of the program?
public class Test { public static void main(String args[]) { int i = 1, j = 0; switch(i) { case 2: j += 6; case 4: j += 1; default: j += 2; case 0: j += 4; } System.out.println("j = " + j); } }
Options- A. j = 0
- B. j = 2
- C. j = 4
- D. j = 6 Discuss
Correct Answer: j = 6
Explanation:
Because there are no break statements, the program gets to the default case and adds 2 to j, then goes to case 0 and adds 4 to the new j. The result is j = 6.- 10. What will be the output of the program?
public class Switch2 { final static short x = 2; public static int y = 0; public static void main(String [] args) { for (int z=0; z < 3; z++) { switch (z) { case x: System.out.print("0 "); case x-1: System.out.print("1 "); case x-2: System.out.print("2 "); } } } }
Options- A. 0 1 2
- B. 0 1 2 1 2 2
- C. 2 1 0 1 0 0
- D. 2 1 2 0 1 2 Discuss
Correct Answer: 2 1 2 0 1 2
Explanation:
The case expressions are all legal because x is marked final, which means the expressions can be evaluated at compile time. In the first iteration of the for loop case x-2 matches, so 2 is printed. In the second iteration, x-1 is matched so 1 and 2 are printed (remember, once a match is found all remaining statements are executed until a break statement is encountered). In the third iteration, x is matched. So 0 1 and 2 are printed.
Comments
There are no comments.
Programming
Copyright ©CuriousTab. All rights reserved.
- 1. What will be the output of the program?