Discussion
Home ‣ Java Programming ‣ Flow Control Comments
- Question
What will be the output of the program?
public class Test { public static void main(String [] args) { int I = 1; do while ( I < 1 ) System.out.print("I is " + I); while ( I > 1 ) ; } }
Options- A. I is 1
- B. I is 1 I is 1
- C. No output is produced.
- D. Compilation error
- Correct Answer
- No output is produced.
ExplanationThere are two different looping constructs in this problem. The first is a do-while loop and the second is a while loop, nested inside the do-while. The body of the do-while is only a single statement-brackets are not needed. You are assured that the while expression will be evaluated at least once, followed by an evaluation of the do-while expression. Both expressions are false and no output is produced.
Flow Control problems
Search Results
- 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. 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.- 3. 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.- 4. 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).- 5. 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.- 6. 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.- 7. What will be the output of the program?
Float f = new Float("12"); switch (f) { case 12: System.out.println("Twelve"); case 0: System.out.println("Zero"); default: System.out.println("Default"); }
Options- A. Zero
- B. Twelve
- C. Default
- D. Compilation fails Discuss
Correct Answer: Compilation fails
Explanation:
The switch statement can only be supported by integers or variables more "narrow" than an integer i.e. byte, char, short. Here a Float wrapper object is used and so the compilation fails.- 8. What will be the output of the program?
int I = 0; outer: while (true) { I++; inner: for (int j = 0; j < 10; j++) { I += j; if (j == 3) continue inner; break outer; } continue outer; } System.out.println(I);
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
Correct Answer: 1
Explanation:
The program flows as follows: I will be incremented after the while loop is entered, then I will be incremented (by zero) when the for loop is entered. The if statement evaluates to false, and the continue statement is never reached. The break statement tells the JVM to break out of the outer loop, at which point I is printed and the fragment is done.- 9. What will be the output of the program?
int x = l, y = 6; while (y--) { x++; } System.out.println("x = " + x +" y = " + y);
Options- A. x = 6 y = 0
- B. x = 7 y = 0
- C. x = 6 y = -1
- D. Compilation fails. Discuss
Correct Answer: Compilation fails.
Explanation:
Compilation fails because the while loop demands a boolean argument for it's looping condition, but in the code, it's given an int argument.while(true) { //insert code here }
- 10. What will be the output of the program?
int x = 3; int y = 1; if (x = y) /* Line 3 */ { System.out.println("x =" + x); }
Options- A. x = 1
- B. x = 3
- C. Compilation fails.
- D. The code runs with no output. Discuss
Correct Answer: Compilation fails.
Explanation:
Line 3 uses an assignment as opposed to comparison. Because of this, the if statement receives an integer value instead of a boolean. And so the compilation fails.
Comments
There are no comments.
Programming
Copyright ©CuriousTab. All rights reserved.
- 1. What will be the output of the program?