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 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 but never definitely assigned before use. Although the assignment s = br.readLine() appears in the while condition, Java's definite assignment analysis requires that local variables be assigned before use in a way that can be verified at compile time. The structure of the try-catch block and the while loop creates a scenario where the compiler cannot guarantee s will be assigned, resulting in a compilation error.

Multiple choice technology web technology

Given: 3. public class Wind { 4. public static void main(String[] args) { 5. foreach: 6. for(int j=0; j<5; j++) { 7. for(int k=0; k< 3; k++) { 8. System.out.print(" " + j); 9. if(j==3 && k==1) break foreach; 10. if(j==0 || j==2) break; 11. } 12. } 13. } 14. } What is the result?

  1. 0 1 2 3

  2. 1 1 1 3 3

  3. 0 1 1 1 2 3 3

  4. 1 1 1 3 3 4 4 4

  5. 0 1 1 1 2 3 3 4 4 4

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

To solve this question, the user needs to understand how the nested for loops work and the use of break statements inside loops.

The code starts with a nested for loop. The outer loop iterates over values of j from 0 to 4, while the inner loop iterates over values of k from 0 to 2. Inside the inner loop, the code prints the value of j and checks two conditions using if statements:

  • If j is equal to 3 and k is equal to 1, the code executes a break statement with the label "foreach", which causes the program to exit both loops and continue executing after line 13.
  • If j is equal to 0 or 2, the code executes a break statement, which only exits the inner loop and continues executing from line 12.

Now, let's go through each option and find the correct answer:

A. 0 1 2 3:

This option is incorrect. The inner loop will only execute when j is equal to 1 or 3, and k is less than 3. When j is equal to 3 and k is equal to 1, the program exits both loops and does not print any more values. Thus, the correct output should not include 4.

B. 1 1 1 3 3:

This option is incorrect. The inner loop will only execute when j is equal to 1 or 3, and k is less than 3. When j is equal to 3 and k is equal to 1, the program exits both loops and does not print any more values. Thus, the correct output should not include 1 or 2.

C. 0 1 1 1 2 3 3:

This option is correct. The inner loop will only execute when j is equal to 1 or 3, and k is less than 3. When j is equal to 3 and k is equal to 1, the program exits both loops and does not print any more values. Thus, the correct output is 0 1 1 1 2 3 3.

D. 1 1 1 3 3 4 4 4:

This option is incorrect. The program never prints the value 4, since the outer loop only iterates from 0 to 4. When j is equal to 3 and k is equal to 1, the program exits both loops and does not print any more values. Thus, the correct output should not include 1 or 2.

E. 0 1 1 1 2 3 3 4 4 4:

This option is incorrect. The program never prints the value 4, since the outer loop only iterates from 0 to 4. When j is equal to 3 and k is equal to 1, the program exits both loops and does not print any more values. Thus, the correct output should not include 1, 2, or 4.

F. Compilation fails:

This option is incorrect. The code compiles without errors.

Therefore, the correct answer is:

The answer is: C. 0 1 1 1 2 3 3.

Multiple choice technology web technology

Given: 3. public class TestDays { 4. public enum Days { MON, TUE, WED }; 5. public static void main(String[] args) { 6. for(Days d : Days.values() ) 7. ; 8. Days [] d2 = Days.values(); 9. System.out.println(d2[2]); 10. } 11. } What is the result? (Choose all that apply.)

  1. TUE

  2. Wed

  3. The output is unpredictable

  4. Compilation fails due to an error on line 4

  5. Compilation fails due to an error on line 6

  6. Compilation fails due to an error on line 8
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Days.values() returns an array of enum constants in declaration order: [MON, TUE, WED]. Array index 2 accesses the third element, which is WED. The empty for-each loop on line 6-7 is valid but does nothing. The code compiles and runs successfully, outputting 'WED'.

Multiple choice technology web technology

What is the output of the following StringBuffer sb1 = new StringBuffer("Amit"); StringBuffer sb2= new StringBuffer("Amit"); String ss1 = "Amit"; System.out.println(sb1==sb2); System.out.println(sb1.equals(sb2)); System.out.println(sb1.equals(ss1)); System.out.println("Poddar".substring(3));

  1. false false false dar

  2. false true false Poddar

  3. Compiler Error

  4. true true false dar
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

sb1 and sb2 are different StringBuffer objects, so sb1==sb2 returns false. StringBuffer's equals() method is not overridden (it inherits from Object), so it compares references - sb1.equals(sb2) also returns false. StringBuffer.equals(String) always returns false due to type mismatch. 'Poddar'.substring(3) returns 'dar' (from index 3 to end). Output: false false false dar.

Multiple choice technology web technology

What is the output (Assuming written inside main) String s1 = new String("amit"); String s2 = s1.replace('m','i'); s1.concat("Poddar"); System.out.println(s1); System.out.println((s1+s2).charAt(5));

  1. Compile error

  2. amitPoddar O

  3. amitPoddar I

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

String objects in Java are immutable - methods like concat() return a new String without modifying the original. After s1.concat('Poddar'), s1 remains 'amit'. s2 becomes 'aiit' after replacing 'm' with 'i'. Concatenating s1+s2 gives 'amitaiit', and charAt(5) returns 'i' (0-indexed: a-m-i-t-a-i, position 5 is 'i').

Multiple choice technology web technology

What is the output of following (Assuming written inside main) String s1 = "Amit"; String s2 = "Amit"; String s3 = new String("abcd"); String s4 = new String("abcd"); System.out.println(s1.equals(s2)); System.out.println((s1==s2)); System.out.println(s3.equals(s4)); System.out.println((s3==s4));

  1. True true true false

  2. True true true true

  3. True false true false

  4. False false true false
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

String literals with the same value reference the same object (s1 == s2 is true), while new String() creates distinct objects (s3 == s4 is false). The equals() method compares content, so all equals calls return true regardless of how the objects were created.

Multiple choice technology programming languages

In the following class definition, which is the first line (if any) that causes a compilation error. Select the one correct answer.public class test { public static void main(String args[]) { char c; int i; c = 'A'; // 1 i = c; //2 c = i + 1; //3 c++; //4 }}

  1. The line labeled 1.

  2. The line labeled 2.

  3. The line labeled 3.

  4. All the lines are correct and the program compiles.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In line 3, adding the integer 1 to the integer 'i' produces an int result. Assigning an int value to a char variable requires an explicit cast to prevent a compiler error from a potential loss of precision.

Multiple choice technology programming languages

What gets printed when the following program is compiled and run. protected class example {public static void main(String args[]) {String test = "abc";test = test + test;System.out.println(test);}}

  1. The program does not compile because statement "test = test + test" is illegal.

  2. The program prints "abc"

  3. The program prints "abcabc"

  4. The class does not compile because the top level class cannot be protected.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Java outer classes cannot be declared with the protected access modifier; they can only be public or package-private. This access modifier violation prevents compilation, making details like string concatenation irrelevant.

Multiple choice technology programming languages

What gets printed when the following program is compiled and run. Select the one correct answer class B { public static void main(String[] args) { int a=10; if(a=10) System.out.println("tomcat"); else System.out.println("root"); } }

  1. tomcat

  2. root

  3. Compile time Error

  4. Run Time Error
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Java requires the condition in an if statement to evaluate to a boolean. The expression 'a=10' is an assignment that returns the integer value 10, resulting in a compile-time type mismatch error instead of evaluating to a truth value.

Multiple choice technology programming languages

import java.lang.*; class A { public static void main(String[] args) { String s1=new String("TCS"); String s2="TCS"; System.out.println(s1==s2); System.out.println(s1.equals(s2)); } } What is the Result?

  1. Compile time error

  2. false true

  3. true false

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

The == operator compares object references in Java, returning false because s1 (heap object) and s2 (string literal) are different objects. The equals() method compares string contents, returning true because both contain 'TCS'. This is a fundamental Java concept distinguishing reference equality from value equality.

Multiple choice technology programming languages

import java.lang.*; class A { public static void main(String[] args) { String s1=new String("TCS"); String s2=s1; System.out.println(s1==s2); System.out.println(s1.equals(s2)); } } What is the Result?

  1. true true

  2. true false

  3. false true

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

When s2 = s1 assigns the reference, both variables point to the same String object in memory. The == operator returns true because they're the same reference. The equals() method also returns true because String's implementation compares content, which is identical.

Multiple choice technology programming languages

import java.lang.*; class A { public static void main(String[] args) { StringBuilder s1="TCS"; String s2=new String("TCS"); System.out.println(s1==s2); System.out.println(s1.equals(s2)); } } What is the Result?

  1. true false

  2. false true

  3. Compile Error

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

This code fails to compile because StringBuilder cannot be initialized with a String literal using the = operator. StringBuilder s1 = 'TCS' is a type mismatch - StringBuilder requires explicit construction via new StringBuilder(). The code won't reach runtime.