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- Question
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
- Correct Answer
- j = 6
ExplanationBecause 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.
Flow Control problems
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- 1. 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.- 2. 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.- 3. 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.- 4. 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 Discuss
Correct Answer: i = 5 and j = 6
Explanation:
This 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.
- 5. 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.- 6. 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.- 7. 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 < 4; z++) { switch (z) { case x: System.out.print("0 "); default: System.out.print("def "); case x-1: System.out.print("1 "); break; case x-2: System.out.print("2 "); } } } }
Options- A. 0 def 1
- B. 2 1 0 def 1
- C. 2 1 0 def def
- D. 2 1 0 def 1 def 1 Discuss
Correct Answer: 2 1 0 def 1 def 1
Explanation:
When z == 0 , case x-2 is matched. When z == 1, case x-1 is matched and then the break occurs. When z == 2, case x, then default, then x-1 are all matched. When z == 3, default, then x-1 are matched. The rules for default are that it will fall through from above like any other case (for instance when z == 2), and that it will match when no other cases match (for instance when z==3).- 8. What will be the output of the program?
int i = 0, j = 5; tp: for (;;) { i++; for (;;) { if(i > --j) { break tp; } } System.out.println("i =" + i + ", j = " + j);
Options- A. i = 1, j = 0
- B. i = 1, j = 4
- C. i = 3, j = 4
- D. Compilation fails. Discuss
Correct Answer: Compilation fails.
Explanation:
If you examine the code carefully you will notice a missing curly bracket at the end of the code, this would cause the code to fail.- 9. 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.- 10. 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.
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- 1. What will be the output of the program?