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

Analyze the following code: public class Test { public static void main(String[] args) { double radius; final double PI= 3.15169; double area = radius * radius * PI; System.out.println("Area is " + area); } }

  1. The program has syntax errors because the variable radius is not initialized.

  2. The program has a syntax error because a constant PI is defined inside a method.

  3. The program has no syntax errors but will get a runtime error because radius is not initialized.

  4. The program compiles and runs fine
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To analyze the given code, we can see that it calculates the area of a circle using a radius and a constant value of PI.

Now let's go through each option and explain why it is right or wrong:

A. The program has syntax errors because the variable radius is not initialized.

This option is partially correct. The variable radius is indeed declared but not initialized. Java requires variables to be initialized before they can be used in calculations. Therefore, the program will not compile due to this syntax error.

B. The program has a syntax error because a constant PI is defined inside a method.

This option is also partially correct. The constant PI is indeed defined inside the main method. However, this is not a syntax error. It is a common practice to define constants inside methods in Java. Therefore, this option is incorrect.

C. The program has no syntax errors but will get a runtime error because radius is not initialized.

This option is incorrect. The program has a syntax error because the variable radius is not initialized. Therefore, the program will not run at all, let alone produce a runtime error.

D. The program compiles and runs fine.

This option is incorrect because the program has a syntax error. It will not compile due to the uninitialized variable radius.

Therefore, the correct answer is:

The Answer is: A

Multiple choice technology programming languages

Analyze the following code. public class Test { int x; public Test(String t) { System.out.println("Test"); } public static void main(String[] args) { Test test = new Test(); System.out.println(test.x); } }

  1. The program has a syntax error because System.out.println method cannot be invoked from the constructor.

  2. The program has a syntax error because x has not been initialized.

  3. The program has a syntax error because you cannot create an object from the class that defines the object.

  4. The program has a syntax error because Test does not have a default constructor.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The Test class defines a constructor with a String parameter (Test(String t)), so Java does NOT auto-generate a no-argument constructor. The main method tries to call 'new Test()' which requires a no-argument constructor that doesn't exist. This is a compilation error, not a runtime error or issues with printing.

Multiple choice technology programming languages

Provide a declaration for i that turns this loop into an infinite loop: while( i != i){ }

  1. String i = "abc"

  2. double i = 0.0/0.0

  3. int i = 3/7

  4. int i = 0
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In IEEE floating-point arithmetic, 0.0/0.0 produces NaN (Not a Number), and NaN != NaN always evaluates to true, creating an infinite loop. Integer division (3/7) equals 0, String comparison compares references, and 0 != 0 is false - only NaN creates this condition.

Multiple choice technology programming languages

Is this a legal Java Program? If so, what does it print? \u0070\u0075\u0062\u006c\u0069\u0063\u0020\u0020\u0020\u0020 \u0063\u006c\u0061\u0073\u0073\u0020\u0055\u0067\u006c\u0079 \u007b\u0070\u0075\u0062\u006c\u0069\u0063\u0020\u0020\u0020 \u0020\u0020\u0020\u0020\u0073\u0074\u0061\u0074\u0069\u0063 \u0076\u006f\u0069\u0064\u0020\u006d\u0061\u0069\u006e\u0028 \u0053\u0074\u0072\u0069\u006e\u0067\u005b\u005d\u0020\u0020 \u0020\u0020\u0020\u0020\u0061\u0072\u0067\u0073\u0029\u007b \u0053\u0079\u0073\u0074\u0065\u006d\u002e\u006f\u0075\u0074 \u002e\u0070\u0072\u0069\u006e\u0074\u006c\u006e\u0028\u0020 \u0022\u0048\u0065\u006c\u006c\u006f\u0020\u0077\u0022\u002b \u0022\u006f\u0072\u006c\u0064\u0022\u0029\u003b\u007d\u007d

  1. Its not possible

  2. print some garbage value

  3. compilation error

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

The Unicode escapes decode to valid Java code: a class named 'Ugly' with a main method that concatenates and prints two strings. Java allows flexible spacing, and string concatenation with + is valid. The program compiles and runs successfully.

Multiple choice technology programming languages

What does it print? public class Elementary { public static void main(String[] args) { System.out.println(12345 + 5432l); } }

  1. Compilation error

  2. Runtime error

  3. 17777

  4. 66666
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The expression adds an integer (12345) and a long literal (5432l). Java automatically promotes the int to long and performs the addition, resulting in 17777, which is printed to the console. Option A is incorrect because the code is syntactically valid. Option B is incorrect because there are no runtime errors. Option D is incorrect because it represents simple concatenation, not arithmetic addition.

Multiple choice technology programming languages

What happens when the following code is compiled and run. Select the one correct answer. for(int i = 2; i < 4; i++) for(int j = 2; j < 4; j++) if(i < j) assert i!=j : i;

  1. The class compiles and runs, but does not print anything.

  2. The number 2 gets printed with AssertionError

  3. compile error

  4. The number 3 gets printed with AssertionError
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The nested loops iterate i=2,3 and j=2,3. The condition if(i < j) filters cases: (2,3) and (2,3) where assert i!=j executes. Since 2!=3 and 2!=3 are both true, the assert never fails. When the assert condition is true (i != j), no AssertionError is thrown. Thus code compiles, runs, and prints nothing.

Multiple choice technology programming languages

What is the output for the below code ? public class A { int k; boolean istrue; static int p; public void printValue() { System.out.print(k); System.out.print(istrue); System.out.print(p); } } public class Test{ public static void main(String argv[]){ A a = new A(); a.printValue();

  1. 0 false 0

  2. 0 true 0

  3. 0 0 0

  4. Compile error - static variable must be initialized before use.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve this question, the user needs to know about Java classes, instance variables, static variables, and method invocation.

The given program has two classes: A and Test. The A class defines three instance variables: k (of type int), istrue (of type boolean), and a static variable p (of type int). Class A also defines a method named printValue() that prints the values of k, istrue, and p, in that order.

In the main() method of the Test class, an object of class A is created and its printValue() method is called.

Since int and boolean instance variables are initialized to their default values of 0 and false respectively, and the static variable p is also initialized to 0, the output of the program will be:

A. 0 false 0

Option A is correct because the program will print the values of k (0), istrue (false), and p (0) in that order.

Option B is incorrect because istrue is initialized to false, not true.

Option C is incorrect because k and istrue are both non-static instance variables, and they will be initialized to their default values of 0 and false, respectively. The static variable p will also be initialized to 0.

Option D is incorrect because there is no compile error in the code.

The Answer is: A. 0 false 0

Multiple choice technology programming languages

What is the output for the below code ? 1. public class Test { 2. public static void main(String... args) { 3. int [] index = new int[5]; 4. System.out.println(index instanceof Object); 5. } 6. }

  1. Compilation fails with an error at line 4

  2. Compilation fails with an error at line 3

  3. true

  4. false
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In Java, arrays are objects. Therefore, any array instance is an instance of Object, and the expression index instanceof Object correctly evaluates to true at runtime.

Multiple choice technology programming languages

What is the output for the below code ? public class A { public void printValue(){ System.out.println("A"); } } public class B extends A { public void printValue(){ System.out.println("B"); } } 1. public class Test { 2. public static void main(String... args) { 3. A b = new B(); 4. newValue(b); 5. } 6. public static void newValue(A a){ 7. if(a instanceof B){ 8. ((B)a).printValue(); 9. } 10. } 11.}

  1. A

  2. Compilation fails with an error at line 4

  3. B

  4. Compilation fails with an error at line 8
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This demonstrates Java polymorphism and instanceof operator. A b = new B() creates a B object but assigns it to A reference. The instanceof check (a instanceof B) returns true because the actual object is B. The cast ((B)a) is safe, and B's printValue() method executes, printing 'B'. This is runtime polymorphism in action.

Multiple choice technology programming languages

What is the output for the below code ? 1. public class Test {2. int i=8; 3. int j=9; 4. public static void main(String[] args){ 5. add(); 6. } 7. public static void add(){ 8. int k = i+j; 9. System.out.println(k); 10. } 11. }

  1. Compilation fails with an error at line 5

  2. Compilation fails with an error at line 8

  3. 17

  4. 0
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The static method add() attempts to access non-static instance variables i and j directly. In Java, static methods cannot access instance variables without an object reference, causing a compilation error at line 8.

Multiple choice technology programming languages

What does it print? public class CleverSwap { public static void main(String[] args) { int x = 1984; int y = 2001; x ^= y ^= x ^= y; System.out.println("x = " + x + "; y = " + y); } }

  1. x = 0; y = 1984

  2. x = 2001; y = 1984

  3. x = 2001; y = 0

  4. x = 1984; y = 2001

  5. x = 0; y = 0

  6. Error
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Java evaluates operands left-to-right. The statement x ^= y ^= x ^= y; is evaluated as x ^= (y ^= (x ^= y)). The initial value of x (1984) is loaded first for the outermost assignment. The innermost x ^= y sets x to 1984 ^ 2001. The middle y ^= x sets y to 2001 ^ (1984 ^ 2001) = 1984. Finally, the outer x ^= y sets x to 1984 ^ 1984 = 0.

Multiple choice technology programming languages

What does it print ? /** * Generated by the IBM IDL-to-Java compiler, version 1.0 * from F:\TestRoot\apps\a1\units\include\PolicyHome.idl * Wednesday, June 17, 1998 6:44:40 o'clock AM GMT+00:00 */ public class Test { public static void main(String[] args) { System.out.print("Hell"); System.out.println("o world"); } }

  1. Hello world

  2. Won't even Compile

  3. Run time error

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The code contains a JavaDoc comment block in the middle of executable code. Java comments cannot appear between method declarations without proper placement - this violates Java syntax rules and prevents compilation.

Multiple choice technology programming languages

What does this program do ? public class BrowserTest { public static void main(String[] args) { System.out.print("iexplore:"); http://www.google.com; System.out.println(":maximize"); } }

  1. Compilation Error

  2. iexplore::maximize

  3. Runtime Error

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The code compiles because 'http:' is treated as a label and '//www.google.com' is a comment. The program prints 'iexplore:' (without newline), then ':maximize' (with newline), resulting in 'iexplore::maximize'. The double colon appears from the printed text and the label.

Multiple choice technology programming languages

What does this program do ? public class Null { public static void greet() { System.out.println("Hello world!"); } public static void main(String[] args) { ((Null) null).greet(); } }

  1. Hello world!

  2. Throw a NullPointerException

  3. Won't even compile

  4. Run time error
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Static methods in Java can be called on a null reference without throwing NullPointerException. The JVM resolves the method call based on the declared type (Null class), not the actual object reference (null). This is a subtle Java behavior often tested in interviews.

Multiple choice technology programming languages

What does it print ? import java.util.*; public class NameGame { public static void main(String args[]) { Map m = new IdentityHashMap(); m.put("Mickey", "Mouse"); m.put("Mickey", "Mantle"); System.out.println(m.size()); } }

  1. Exception

  2. Stack Overflow

  3. 2

  4. 1
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In Java, an IdentityHashMap uses reference equality (==) rather than object equality (equals()) for keys. Since the two string literals "Mickey" are interned, they refer to the exact same string object in memory. Therefore, the second put operation overwrites the first, resulting in a size of 1.