Computer Knowledge

Programming Output Evaluation

1,953 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

package staticinit; public class TestStaticInit { { System.out.print("I am in Init "); } TestStaticInit() { System.out.print("Constructor. "); } static { System.out.print("I am in Static. "); } static { System.out.print("I am in Static1. "); } public static void main(String[] args) { } }

  1. No output

  2. I am in Static. I am in Static1.

  3. Run time exception

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

Static blocks execute when the class is loaded, even if no objects are created. Since main() is static, merely loading the class to execute main() triggers all static blocks in order. Instance initialization blocks and constructors only run when 'new' is used.

Multiple choice technology programming languages

package statictest; class TestStaticSub { static int p = 100; public static void printP() { System.out.println("P in parent class..." + p); } } public class TestStaticVariable extends TestStaticSub { public static void printP() { p = 200; System.out.println("P in sub class ..." + p); } public static void main(String[] args) { TestStaticSub tsv = new TestStaticVariable(); tsv.printP(); } }

  1. P in sub class ... 200

  2. Compiler error

  3. P in parent class...100

  4. Run time exception
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Static methods are resolved at compile‑time based on the reference type. The variable tsv is declared as TestStaticSub, so TestStaticSub.printP() is called, which prints the unchanged static field p (100). The subclass’s method is never invoked, making the stored answer correct.

Multiple choice technology programming languages

package statictest; public class TestStaticVar { static int p =100; static void setP(int p) { this.p = p; System.out.print("p== "+p); } public static void main(String[] args) { setP(500); System.out.print("p=="+p); } }

  1. p== 500 p== 100

  2. Run time exception

  3. Compiler error

  4. p== 500 p== 500
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

You cannot use 'this' inside a static method because 'this' refers to the current instance, but static methods belong to the class not to any instance. The compiler will catch this error at compile-time. The static variable 'p' should be accessed directly or via the class name, not through 'this'.

Multiple choice technology programming languages

public class test { Line 1. public static void main(String [] a) { Line 2. assert a.length == 1; Line 3. } Line 4. } Which two will produce an AssertionError? (Choose two.) A. java test B. java -ea test C. java test file1 D. java -ea test file1 E. java -ea test file1 file2 F. java -ea:test test file1

  1. AB

  2. BC

  3. AE

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

Running with -ea enables assertions. If no arguments are passed (length 0) or more than one argument is passed (length 2 or more), the assertion a.length == 1 fails, raising an AssertionError.

Multiple choice technology programming languages

package testing; public class SimpleClone implements Cloneable { int number = 100; int[] intArray = new int[10]; public SimpleClone() { int number = 100; for (int i = 0; i < intArray.length; i++) { intArray[i] = number; number++; } } public static void main(String[] args) { SimpleClone testclone1 = new SimpleClone(); SimpleClone testclone2; try { testclone2 = (SimpleClone) testclone1.clone(); testclone1.intArray[0] = 1200; testclone2.intArray[0] = 1400; testclone1.number = 1200; System.out.print(" testclone1 intArray[0] = " + testclone1.intArray[0]); System.out.print(" testclone1one2 intArray[0] = " + testclone2.intArray[0]); System.out.print(" testclone1 value of number = " + testclone1.number); System.out.print(" testclone2 value of number = " + testclone2.number); } catch (CloneNotSupportedException e) { // TODO Auto-generated catch block e.printStackTrace(); } } }

  1. testclone1 intArray[0] = 1400 testclone1one2 intArray[0] = 1400 testclone1 value of number = 1200 testclone2 value of number = 100

  2. testclone1 intArray[0] = 1200 testclone1one2 intArray[0] = 1400 testclone1 value of number = 1200 testclone2 value of number = 100

  3. testclone1 intArray[0] = 1400 testclone1one2 intArray[0] = 1400 testclone1 value of number = 1200 testclone2 value of number = 1200

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

Object.clone() performs shallow copy. Primitive 'number' is copied (testclone1.number=1200 doesn't affect testclone2.number which stays 100). But 'intArray' is a reference, so both objects share the same array. When testclone2.intArray[0]=1400 executes, it overwrites the earlier testclone1.intArray[0]=1200 assignment. Both objects now have intArray[0]=1400 because they reference the same array.

Multiple choice technology programming languages

package io; import java.io.*; public class TestFile { public static void main(String[] args) { File file = new File("test_file.txt"); System.out.print(" Fle Name = " + file.getName()); file.delete(); file.renameTo(new File("test_file2.txt")); System.out.print(" Fle Name = " + file.getName()); } }

  1. Null pointer Exception

  2. Fle Name = test_file.txt Fle Name = test_file.txt

  3. Compiler Error

  4. Fle Name = test_file.txt Fle Name = test_file2.txt
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The File object maintains its reference regardless of file system state. file.getName() returns 'test_file.txt' initially. After file.delete(), the file is removed but the File object still holds the original name. The renameTo() operation fails (returns false) because the file was already deleted, so getName() still returns 'test_file.txt'. The File object's state doesn't automatically sync with the filesystem.

Multiple choice technology programming languages

package io; import java.io.*; public class TestFile1 { public static void main(String[] args) { //No file exists by the name test_file.txt File file = new File("test_file.txt"); System.out.print(" Fle Name = "+file.getName()); System.out.print(" File Status = "+file.exists()); } }

  1. Fle Name = test_file.txt File Status = false

  2. Compiler Error

  3. Null Pointer Exception

  4. Fle Name = test_file.txt File Status = true
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The File constructor in Java does not create a file on the disk or check for its existence; it merely creates an in-memory representation of the path. Thus, getName() returns the file name string "test_file.txt", and exists() returns false.

Multiple choice technology programming languages

Integer division by zero throws an exception.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Integer division by zero (e.g., 5/0) throws ArithmeticException, unlike floating-point division by zero (e.g., 5.0/0) which produces Infinity. This is because integer arithmetic cannot represent infinity - the operation is undefined for integers and must signal an error.

Multiple choice technology programming languages

package simplejava;public class TestConstructor{ int p ; public void TestConstructor(int p) { this.p=p; } public static void main(String[] args) { TestConstructor tp = new TestConstructor(10); TestConstructor tp = new TestConstructor(); System.out.println("p = "+tp.p); }}

  1. Compiler Error

  2. Exception

  3. p = 0

  4. p = 10
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The method TestConstructor has a void return type, making it a regular method instead of a constructor. Since no valid constructor is defined, Java will provide a default no-arg constructor. However, the code attempts to call TestConstructor(10) which doesn't exist as a constructor.

Multiple choice technology web technology

package simplejava; public class TestConstructor1 { int p; TestConstructor1() { System.out.println(&quot;I am in Constructor&quot;); } TestConstructor1(int p) { this.p =p; System.out.println(&quot;p = &quot;+p); this(); } public static void main(String[] args) { TestConstructor1 tp1 = new TestConstructor1(10); } }

  1. p = 10

  2. p = 0

  3. Compiler Error

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

In Java, a constructor call (this()) must be the first statement in a constructor. Here, the parameterized constructor attempts to assign this.p=p and print before calling this(), which violates this rule. The compiler enforces that constructor chaining happens first.

Multiple choice technology programming languages

package simplejava; public class TestConstructor2 { int p; TestConstructor2() { System.out.print(" I am in Constructor"); } TestConstructor2(int p) { this(); this.p = p; System.out.print(" p = " + p); } public static void main(String[] args) { TestConstructor2 tp1 = new TestConstructor2(10); } }

  1. I am in Constructor p = 10

  2. I am in Constructor p = 0

  3. Compiler Error

  4. Exception
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The parameterized constructor correctly calls this() as its first statement, executing the no-arg constructor first. It prints 'I am in Constructor', then assigns p and prints 'p = 10'. The output is the concatenation of both print statements.

Multiple choice technology programming languages

package simplejava;public class TestMain { TestMain() { System.out.println("I am in Constructor"); } TestMain(String message) { System.out.println("Message " + message); } public static void main(String[] args) { TestMain tm = new TestMain(); } public static void main(String args) { TestMain tm = new TestMain("Hello"); }}

  1. Message Hello

  2. I am in Constructor

  3. Compiler Error

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

The class has two overloaded main methods - one with String[] args and one with String args. When the JVM launches, it calls the standard main(String[] args) which invokes the no-arg constructor, printing 'I am in Constructor'. The second main method is never called.

Multiple choice technology programming languages

package simplejava;public class TestMain1 { TestMain1() { System.out.println("I am in Constructor"); } TestMain1(String message) { System.out.println("Message " + message); } public void main(String[] args) { TestMain1 nm = new TestMain1(); TestMain1 nm1 = new TestMain1("Hello"); }}

  1. Compiler Error

  2. Exception

  3. I am in Constructor

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

The main method is not declared static, which is required by the JVM. When you try to run this class, the JVM cannot find a valid entry point and will throw an exception at runtime stating that the main method must be static.

Multiple choice technology web technology

Given: 3. import java.io.*; 4. public class ReadingFor { 5. public static void main(String[] args) { 6. String s; 7. try { 8. FileReader fr = new FileReader("myfile.txt"); 9. BufferedReader br = new BufferedReader(fr); 10. while((s = br.readLine()) != null) 11. System.out.println(s); 12. br.flush(); 13. } catch (IOException e) { System.out.println("io error"); } 16. } 17. } And given that myfile.txt contains the following two lines of data: ab cd What is the result?

  1. ab

  2. abcd

  3. ab cd

  4. a b c D

  5. Compilation fails
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

The variable s is declared inside the try block but used in the while condition. However, s is not properly initialized or assigned before use. More critically, br.flush() is called on BufferedReader which doesn't have a flush() method - only the underlying Writer does. This causes compilation failure.

Multiple choice technology web technology

Given: public static void main(String[] args) { // INSERT DECLARATION HERE for (int i = 0; i <= 10; i++) { List row = new ArrayList(); for (int j = 0; j <= 10; j++) row.add(i * j); table.add(row); } for (List row : table) System.out.println(row); } Which statements could be inserted at // INSERT DECLARATION HERE to allow this code to compile and run? (Choose all that apply.)

  1. List<List<Integer>> table = new List<List<Integer>>();

  2. List<List<Integer>> table = new ArrayList<List<Integer>>();

  3. List<List<Integer>> table = new ArrayList<ArrayList<Integer>>();

  4. List<List, Integer> table = new List<List, Integer>();

  5. List<List, Integer> table = new ArrayList<List, Integer>();

  6. List<List, Integer> table = new ArrayList<ArrayList, Integer>();
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The code creates a 2D table using nested loops, so table must be a List of List. Option B correctly declares List> table = new ArrayList>() - ArrayList implements List, and the generic type matches. Option A fails because List is an interface and cannot be instantiated. Option C fails because ArrayList> cannot be assigned to List> - the inner generic type must match exactly. Options D-F use invalid syntax with commas instead of proper generic notation.