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 web technology

public class Name { private String first, last; public Name(String first, String last) { this.first = first; this.last = last; } public boolean equals(Object o) { if (!(o instanceof Name)) return false; Name n = (Name)o; return n.first.equals(first) && n.last.equals(last); } public static void main(String[] args) { Set s = new HashSet(); s.add(new Name("Donald", "Duck")); System.out.println(s.contains(new Name("Donald", "Duck"))); } } What is the output?

  1. true

  2. false

  3. Compile Error

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice technology programming languages

public class Change { public static void main(String args[]) { System.out.println(2.00 - 1.10); } } What is the output

  1. 0.9

  2. 0.90

  3. It varies

  4. None of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Due to binary floating-point representation, 2.00 - 1.10 does not equal 0.9 exactly. In binary, 0.1 cannot be represented precisely (it's a repeating fraction in binary). The actual result is 0.8999999999999999. Since none of the options A (0.9), B (0.90), or C (It varies - suggests randomness) are correct, the answer is D (None of the above).

Multiple choice technology programming languages

public class test08 { public static void main(String[] args) { final int start = Integer.MAX_VALUE - 100; final int end = Integer.MAX_VALUE; int count = 0; for (int i = start; i <= end; i++) count++; System.out.println(count); } } What will be the output

  1. 100

  2. 101

  3. 102

  4. None of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This loop appears to count from (Integer.MAX_VALUE - 100) to Integer.MAX_VALUE, which should be 101 iterations. However, when i equals Integer.MAX_VALUE and we execute i++, it overflows to Integer.MIN_VALUE (most negative int). The condition i <= end is then always true (since MIN_VALUE <= MAX_VALUE), creating an infinite loop. The count will increase indefinitely.

Multiple choice technology programming languages

public class test08 { public static void main(String[] args) { System.out.print("H" + "a"); System.out.print('H' + 'a'); } } What is the output?

  1. HaHa

  2. Ha

  3. Compile error

  4. None of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The first print uses string concatenation: "H" + "a" = "Ha". The second print involves character arithmetic: 'H' + 'a' promotes chars to int, resulting in 72 + 97 = 169. The output is "Ha169" (or "Ha" followed by 169). Since this exact output isn't listed, the answer is "None of the above".

Multiple choice technology programming languages

public class test08 { public test08(Object o) { System.out.println("Object"); } public test08(double[] dArray) { System.out.println("double array"); } public static void main(String[] args) { new test08(null); } }

  1. Object

  2. double array

  3. Compile Error

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

When null is passed to overloaded constructors, Java selects the most specific type. Since double[] is more specific than Object, the compiler selects the double array constructor. This follows Java's overload resolution rules where reference type arrays are more specific than Object.

Multiple choice technology programming languages

#include class NewInt { int num; public: NewInt(int n=0):num(n) {printf("cons\n");} int getInt() const {return num;} void setInt(int n){num =n;} ~NewInt(){printf("Dest\n");} }; int main() { NewInt a[5]; for(int i=0;i<=5;i++) a[i].setInt(i+10); for(int i=0;i<=5;i++) printf("%d\n",a[i].getInt()); return 0; } What does the above code snippet do?

  1. It creates an array of classes

  2. Gives compile time error

  3. It creates an array of objects

  4. It does nothing
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice technology programming languages

What happens when the following code is compiled and run. Select the one correct answer. for(int i = 1; i < 3; i++) for(int j = 3; j >= 1; j--) assert i!=j : i;

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

  2. The number 1 gets printed with AssertionError

  3. The number 2 gets printed with AssertionError

  4. The number 3 gets printed with AssertionError

  5. The program generates a compilation error.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In the first iteration, i = 1. In the inner loop, when j decreases to 1, the assertion condition i != j becomes 1 != 1 (false). Since assertions are enabled, this triggers an AssertionError with the value of i (which is 1). Other options are wrong because the assertion fails before i can reach 2.

Multiple choice technology programming languages
  1. interface I{ void f1(); // 1 public void f2(); // 2 protected void f3(); // 3 private void f4(); // 4 } which lines generate compile time errors?

  1. compiletime error at lines 1,2,3,4

  2. compiletime error at line 3

  3. compiletime error at line 1

  4. compiletime error at lines 3,4

  5. None of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In standard Java, interface methods are implicitly public and abstract. Explicitly declaring them as protected (line 3) or private (line 4) causes compile-time errors. Distractors are incorrect because lines 1 and 2 are valid method declarations that compile without issues.

Multiple choice technology programming languages

class C1 { static interface I { static class C2 { } } public static void main(String a[]) { C1.I.C2 ob1=new C1.I.C2(); System.out.println("object created"); } } What is the result of attempting to compile and run the program?

  1. prints object created

  2. Compile time error

  3. Runtime Excepion

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Java allows static nested classes inside interfaces. The syntax C1.I.C2 is valid - C2 is a static class inside interface I, which is inside class C1. The code compiles successfully and creates the object, printing 'object created'. No compile-time or runtime errors occur.

Multiple choice technology programming languages

You have the following code in a file called Test.java class Base{ public static void main(String[] args){ System.out.println("Hello"); } } public class Test extends Base{} What will happen if you try to compile and run this?

  1. It will fail to compile.

  2. Runtime error

  3. Compiles and runs with no output.

  4. Compiles and runs printing "Hello"
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The code compiles and runs successfully because the main() method is inherited from the Base class. Even though Test extends Base, static methods are not inherited in the traditional sense - they belong to the class where they're defined. However, when you run 'java Test', it executes Base.main() which prints "Hello".

Multiple choice technology programming languages

What is the result of trying to compile and run this program. public class Test{ public static void main(String[] args){ int[] a = {1}; Test t = new Test(); t.increment(a); System.out.println(a[a.length - 1]); } void increment(int[] i){ i[i.length - 1]++; } }

  1. Compiler error.

  2. Compiles and runs printing out 2

  3. Compiles and runs printing out 1

  4. An ArrayIndexOutOfBounds Exception at runtime
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Java passes references by value. The array reference a is copied to parameter i in the increment method, allowing modification of the original array's elements. The method correctly increments the last element from 1 to 2. The other options are wrong because the code compiles successfully and does not throw runtime exceptions.

Multiple choice technology programming languages

What will happen if you try to compile and run this ? public class Test{ static{ print(10); } static void print(int x){ System.out.println(x); System.exit(0); } }

  1. Compiler error.

  2. Will throw a NoSuchMethod error at runtime.

  3. It will compile and run printing out "10"

  4. It will run with no output.

  5. It will run and print "10" and then crash with an error.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The code compiles and runs successfully. Static blocks are executed when the class is loaded, and static methods can be called from static blocks. The print(10) method is static, so it's accessible from the static block. The program prints "10" and then exits immediately due to System.exit(0).