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- Question
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.
- Correct Answer
- Compilation fails.
ExplanationLine 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.
Flow Control problems
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- 1. 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 }
- 2. 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.- 3. 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.- 4. 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.- 5. 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 Discuss
Correct Answer: No output is produced.
Explanation:
There 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.- 6. What will be the output of the program?
public class SwitchTest { public static void main(String[] args) { System.out.println("value =" + switchIt(4)); } public static int switchIt(int x) { int j = 1; switch (x) { case l: j++; case 2: j++; case 3: j++; case 4: j++; case 5: j++; default: j++; } return j + x; } }
Options- A. value = 2
- B. value = 4
- C. value = 6
- D. value = 8 Discuss
Correct Answer: value = 8
Explanation:
Because there are no break statements, once the desired result is found, the program continues though each of the remaining options.- 7. 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
- 8. What will be the output of the program?
for (int i = 0; i < 4; i += 2) { System.out.print(i + " "); } System.out.println(i); /* Line 5 */
Options- A. 0 2 4
- B. 0 2 4 5
- C. 0 1 2 3 4
- D. Compilation fails. Discuss
Correct Answer: Compilation fails.
Explanation:
Compilation fails on the line 5 - System.out.println(i); as the variable i has only been declared within the for loop. It is not a recognised variable outside the code block of loop.- 9. What will be the output of the program?
public class If1 { static boolean b; public static void main(String [] args) { short hand = 42; if ( hand < 50 && !b ) /* Line 7 */ hand++; if ( hand > 50 ); /* Line 9 */ else if ( hand > 40 ) { hand += 7; hand++; } else --hand; System.out.println(hand); } }
Options- A. 41
- B. 42
- C. 50
- D. 51 Discuss
Correct Answer: 51
Explanation:
In Java, boolean instance variables are initialized to false, so the if test on line 7 is true and hand is incremented. Line 9 is legal syntax, a do nothing statement. The else-if is true so hand has 7 added to it and is then incremented.- 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 y: System.out.print("0 "); /* Line 11 */ case x-1: System.out.print("1 "); /* Line 12 */ case x: System.out.print("2 "); /* Line 13 */ } } } }
Options- A. 0 1 2
- B. 0 1 2 1 2 2
- C. Compilation fails at line 11.
- D. Compilation fails at line 12. Discuss
Correct Answer: Compilation fails at line 11.
Explanation:
Case expressions must be constant expressions. Since x is marked final, lines 12 and 13 are legal; however y is not a final so the compiler will fail at line 11.
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- 1. What will be the output of the program?