Computer Knowledge

Programming Output Evaluation

1,721 Questions

Programming output evaluation tests the ability to trace code execution in languages like C, Java, and SAS. It focuses on arrays, loops, pointers, and data type conversions. These technical questions are standard in computer knowledge sections for IT officer and bank exams.

Java string bufferC language pointersLoop execution outputsData type conversionsMacro variable evaluation

Programming Output Evaluation Questions

Multiple choice technology programming languages
  1. public class Test { 12. public static void main(String [] args) { 13. int x =5; 14. boolean b1 = true; 15. boolean b2 = false; 16. 17.if((x==4) && !b2) 18. System.out.print(”l “); 19. System.out.print(”2 “); 20. if ((b2 = true) && b1) 21. System.out.print(”3 “); 22. } 23. } What is the result?

  1. 2

  2. 3

  3. 1 2

  4. 2 3

  5. 1 2 3

  6. Compilation fails.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The first condition (x==4) && !b2 is false because x=5, so '1' is not printed. Print('2') always executes, printing '2'. The second condition (b2 = true) assigns true to b2, then evaluates (true && b1) which is true, so '3' is printed. Output: '2 3'.

Multiple choice technology programming languages

Click the Exhibit button. 1. public class Test { 2. int x= 12; 3. public void method(int x) { 4. x+=x; 5. System.out.println(x); 6. } 7. } Given: 34. Test t = new Test(); 35. t.method(5); What is the output from line 5 of the Test class?

  1. 5

  2. 10

  3. 12

  4. 17

  5. 24
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The output from line 5 of the Test class would be 10.

Explanation:

In the Test class, there is an instance variable x with a value of 12.

The method takes a parameter x and performs an addition operation x += x, which is equivalent to x = x + x.

When the method is called with an argument of 5 (t.method(5)), the local variable x inside the method is set to 5.

The addition operation x += x is then performed, resulting in x being updated to 10.

Finally, System.out.println(x) prints the value of x, which is 10.

Therefore, the output from line 5 of the Test class would be 10, so the correct answer is C.

Multiple choice technology programming languages

Given: 11. public class Test { 12. public enum Dogs {collie, harrier, shepherd}; 13. public static void main(String [] args) { 14. Dogs myDog = Dogs.shepherd; 15. switch (myDog) { 16. case collie: 17. System.out.print(”collie “); 18. case default: 19. System.out.print(”retriever “); 20. case harrier: 21. System.out.print(”harrier “); 22. } 23. } 24. } ‘What is the result?

  1. harrier

  2. shepherd

  3. retriever

  4. Compilation fails.

  5. retriever harrier

  6. An exception is thrown at runtime.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The switch statement contains a critical syntax error on line 18: 'case default:' is invalid. In Java switch statements, 'default' is a keyword that stands alone and cannot be combined with 'case'. The correct syntax is simply 'default:' without the word 'case' preceding it. This causes compilation to fail regardless of the enum value being tested.

Multiple choice technology programming languages

Given: 11. public static void main(String[] args) { 12. String str = “null’; 13. if (str == null) { 14. System.out.println(”null”); 15. } else (str.length() == 0) { 16. System.out.println(”zero”); 17. } else { 18. System.out.println(”some”); 19. } 20. } ‘What is the result?

  1. null

  2. zero

  3. some

  4. Compilation fails.

  5. An exception is thrown at runtime.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Line 15 has a syntax error: 'else (str.length() == 0)' is missing the keyword 'if'. In Java, 'else' must be followed immediately by a statement or block, or by 'if' for an else-if clause. The parentheses alone make no sense syntactically. This causes compilation failure regardless of the string value being tested.

Multiple choice technology programming languages

Given: 11. public static void test(String str) { 12. int check = 4; 13. if (check = str.length()) { 14. System.out.print(str.charAt(check -= 1) +“, “); 15. } else { 16. System.out.print(str.charAt(0) + “, “); 17. } 18. } and the invocation: 21. test(”four”); 22. test(”tee”); 23. test(”to”); What is the result?

  1. r, t, t,

  2. r, e, o,

  3. Compilation fails.

  4. An exception is thrown at runtime.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Line 13 attempts to assign an integer (str.length() returns int) to a boolean variable (check is int, but if condition requires boolean). In Java, check = str.length() is an assignment, not a comparison, and the types are incompatible for the if condition.

Multiple choice technology programming languages

Given: 11. public static void test(String str) { 12. int check = 4; 13. if (check = str.length()) { 14. System.out.print(str.charAt(check -= 1) +“, “); 15. } else { 16. System.out.print(str.charAt(0) + “, “); 17. } 18. } and the invocation: 21. test(”four”); 22. test(”tee”); 23. test(”to”); What is the result?

  1. r, t, t,

  2. r, e, o,

  3. Compilation fails.

  4. An exception is thrown at runtime.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Line 13 attempts to assign an integer (str.length() returns int) to a boolean variable (check is int, but if condition requires boolean). In Java, check = str.length() is an assignment, not a comparison, and the types are incompatible for the if condition.

Multiple choice technology programming languages

What is the result of compiling and running this program? class Clove{ void quality(Clove m){ System.out.println("Clove is nice"); } } class Mint extends Clove{ void quality(Mint c){ System.out.println("Mint is ok"); } } class Sage extends Mint{ void quality(Sage h){ System.out.println("Sage is average"); } } public class Test{ public static void main(String[] args){ Clove h = new Sage(); Mint c = new Sage(); c.quality(h); } }

  1. prints "Clove is nice"

  2. prints "Mint is ok"

  3. prints "Sage is average"

  4. Class cast Exception at runtime.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The variable c is declared as type Mint. The method c.quality(h) calls quality() on a Mint reference. The argument h is declared as Clove, so the method signature quality(Mint c) in class Mint doesn't match. JVM looks for quality(Clove) in Mint's hierarchy, finds it in Clove class (parent), and executes that due to method resolution rules.

Multiple choice technology programming languages

What will this program print out ? class Base{ int value = 0; Base(){ addValue(); } void addValue(){ value += 10; } int getValue(){ return value; } } class Derived extends Base{ Derived(){ addValue(); } void addValue(){ value += 20; } } public class Test { public static void main(String[] args){ Base b = new Derived(); System.out.println(b.getValue()); } }

  1. 10

  2. 20

  3. 30

  4. 40
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

During instantiation of Derived, the Base constructor is called first, which invokes addValue(). Because of dynamic binding, the overridden addValue() in Derived runs, adding 20 to value. Then, the Derived constructor runs and calls addValue() again, adding another 20, resulting in 40.

Multiple choice technology programming languages

Given: 1. public class Threads4 { 2. public static void main (String[] args) { 3. new Threads4().go(); 4. } 5. public void go() { 6. Runnable r = new Runnable() { 7. public void run() { 8. System.out.print(”foo”); 9. } 10. }; 11. Thread t = new Thread(r); 12. t.start(); 13. t.start(); 14. } 15. } What is the result?

  1. Compilation fails.

  2. An exception is thrown at runtime.

  3. The code executes normally and prints ‘foo”.

  4. The code executes normally, but nothing is printed.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Calling start() twice on the same Thread object throws IllegalThreadStateException at runtime. Once a thread has been started, subsequent calls to start() cause an exception - you cannot restart a thread.

Multiple choice technology programming languages

Click the Exhibit button. 10. class Inner { 11. private int x; 12. public void setX( int x) { this.x = x; } 13. public int getX() { return x; } 14. } 15. 16. class Outer { 17. private Inner y; 18. public void setY( Inner y) { this.y = y; } 19. public Inner getY() { return y; } 20. } 21. 22. public class Gamma { 23. public static void main( String[] args) { 24. Outer o = new Outer(); 25. Inner i = new Inner(); 26.int n=10; 27. i.setX(n); 28. o.setY(i); 29. // insert code here 30. System.out.println( o.getY().getX()); 31. } 32. } Which three code fragments, added individually at line 29, produce the output 100? (Choose three.)

  1. n = 100;

  2. i.setX( 100);

  3. o.getY().setX( 100);

  4. i = new Inner(); i.setX( 100);

  5. o.setY( i); i = new Inner(); i.setX( 100);

  6. i = new Inner(); i.setX( 100); o.setY( i);
Reveal answer Fill a bubble to check yourself
B,C,F Correct answer
Explanation

The code passes an Inner object to Outer via setY(). Options B, C, and F all work: B modifies i directly before it was passed to o, C accesses the Inner through o and modifies it, and F creates a new Inner, modifies it, then replaces it in o. Options A, D, and E fail because they modify a different object or n (a primitive).

Multiple choice technology programming languages

Given: 1. public class TestString3 { 2. public static void main(String[] args) { 3. // insert code here 5. System.out.println(s); 6. } 7. } Which two code fragments, inserted independently at line 3, generate the output 4247? (Choose two.) A. String s = “123456789”; s = (s-”123”).replace(1,3,”24”) - “89”; B. StringBuffer s = new StringBuffer(”123456789”); s.delete(0,3).replace( 1,3, “24”).delete(4,6); C. StringBuffer s = new StringBuffer(”123456789”); s.substring(3,6).delete( 1 ,3).insert( 1, “24”); D. StringBuilder s = new StringBuilder(”123456789”); s.substring(3,6).delete( 1 ,2).insert( 1, “24”); E. StringBuilder s = new StringBuilder(”123456789”); s.delete(0,3).delete( 1 ,3).delete(2,5).insert( 1, “24”);

  1. A

  2. B

  3. C

  4. D

  5. E
Reveal answer Fill a bubble to check yourself
B,E Correct answer
Explanation

Option B: StringBuffer starts with '123456789', delete(0,3) removes '123' leaving '456789', replace(1,3,'24') affects positions 1-2 ('56') replacing with '24' resulting '424789', delete(4,6) removes '89' leaving '4247'. Option E: StringBuilder delete operations transform '123456789' to '4247'. Other options fail due to method chaining issues or incorrect indices.

Multiple choice technology programming languages

Given: 33. try { 34. // some code here 35. } catch (NullPointerException e1) { 36. System.out.print(”a”); 37. } catch (RuntimeException e2) { 38. System.out.print(”b”); 39. } finally { 40. System.out.print(”c”); 41. } What is the result if a NullPointerException occurs on line 34?

  1. c

  2. a

  3. ab

  4. ac

  5. bc

  6. abc
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When NullPointerException occurs at line 34, it's caught by the first catch block (line 35-36) which prints 'a'. The finally block (line 39-40) always executes, printing 'c'. The output is 'ac'. The second catch block is skipped because the first one matches.

Multiple choice technology programming languages

What all gets printed when the following code is compiled and run? Select the three correct answers. public class xyz { public static void main(String args[]) { for(int i = 0; i < 2; i++) { for(int j = 2; j>= 0; j--) { if(i == j) break; System.out.println("i=" + i + " j="+j); } } } }

  1. i=0 j=0

  2. i=0 j=1

  3. i=0 j=2

  4. i=1 j=0

  5. i=1 j=1

  6. i=1 j=2
Reveal answer Fill a bubble to check yourself
B,C,F Correct answer
Explanation

Tracing the loops: When i=0, j prints 2,1 before break at j=0. When i=1, j prints 2 before break at j=1. When i=2, j prints 2,1,0 but i=2 after loop ends. Output: i=0 j=2, i=0 j=1, i=1 j=2.

Multiple choice technology programming languages

What gets printed when the following program is compiled and run. Select the one correct answer. class test { public static void main(String args[]) { int i,j,k,l=0; k = l++; j = ++k; i = j++; System.out.println(i); } }

  1. 0

  2. 1

  3. 2

  4. 3
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The post-increment l++ assigns the current value 0 to k, then increments l to 1. Pre-increment ++k increments k to 1 first, then assigns 1 to j. Post-increment j++ assigns 1 to i, then increments j to 2. The output is 1.