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

What is the output of the given program?

public class arr
{
public static void main(String[] args)
{
int array[] = {0, 1, 2, 3, 4};
int key = 3;
for (int pos = 0;
pos < array.length; ++pos)
{
if (array[pos] == key)
{
break;
}
}
System.out.print("Found " + key + "at " + pos);
}
}

  1. Found 3 at 2

  2. Found 3 at 3

  3. Found 3 at 4

  4. Compilation fails

  5. Run time error
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The following line does not compile: System.out.print (found + key + at + pos). 

Multiple choice

How many times is 0 printed as part of output?

public class ScopeTest
{
int z;
public static void main(String[] args)
{
String[] table = {aa, bb, cc};
for (String ss: table)
{
int ii = 0;
while (ii < table.length)
{
System.out.println(ss + , + ii);
ii++;
}
}
}
}

  1. Zero

  2. Once

  3. Twice

  4. Thrice

  5. Compilation error
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This is the correct choice. The for statement, for (String ss: table) is executed one time for each of the three elements in table. 

Multiple choice

What is the output of the given code?

Int [] [] array = {{0}, {0, 1}, {0, 2, 4}, {0, 3, 6, 9}, {0, 4, 8, 12, 16}};
Systemout.printIn(array [4] [1]);
System.out.printIn (array) [1][4]);

  1. 4 null

  2. Null 4

  3. An exception is thrown at run time.

  4. 4 an ArrayIndexOutofbound exception is thrown.

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

This is the correct answer. The first print in statement, system.out.println(array [4][1]), works fine. It selects the element/array with index 4, {0, 4, 8, 12, 16} and from this array, it selects the element with index 1, which is 4. So, 4 gets printed. The second print in statement, system.out.println(array) [1][4]) fails. It selects the array/element with index 1,{0, 1} and from this array, it tries to select the element with index 4. This causes an exception. So, an exception in thread main java.lang.ArrayIndexOutOfBoundsException occurs.

Multiple choice

What is the output of the given program?

public class ScopeTest
{
int z;
public static void main(String[] args)
{
ScopeTest myScope = new ScopeTest();
int z = 6;
System.out.println(z);
myScope.doStuff();
System.out.println(z);
System.out.println(myScope.z);
}
void doStuff()
{
int z = 5;
doStuff2();
System.out.println(z);
}
void doStuff2()
{
z = 4;
}
}

  1. 6 5 6 4

  2. 6 5 5 4

  3. 6 5 6 6

  4. Run time error

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

Within main program, z is assigned value 6. So, z = 6 is printed first. Within doStuff, z is assigned 5. DoStuff2 locally sets z to 4 (but MyScope.z is set to 4), but in dostuff, z is still 5. So, z=5 is printed. Again z is printed within main (with local z set to 6). So, z = 6 is again printed and finally myScope.z is printed. 

Multiple choice

What is the output of the given program?

public class ScopeTest
{
public static void main (String [] args)
{
int[][] xx[][][] = null;
System.out.println(xx);
}
}

  1. Compilation Fails

  2. Null

  3. 0

  4. Java.lang.ArrayIndexOutOfBoundException

  5. Java.lang.NullPointerException
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This is the correct choice because an array variable (here xx) can very well have the null value. Null is the reserved constant used in Java to represent a void reference, i.e. a pointer to nothing. Internally, it is just a binary 0, but in the high level Java language, it is a magic constant, quite distinct from zero, that internally could have any representation. So, this answer is correct.

Multiple choice

What is the output of the given program?

public class ScopeTest
{
public static void main (String [] args)
{
int ii = 2;
do
{
System.out.println (ii);
}
while (--ii);
}
}

  1. 1

  2. 2

  3. Null

  4. Infinite loop

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

This is the correct choice because the line while (--ii); will cause the compilation to fail. --ii is not a boolean value. A correct line would be while (--ii>0); we can pass only Boolean values to the while loop. So, this is the correct choice.

Multiple choice

Which of the following is the output of the given program?

public static void main(Stirng[] args)
{
int x = 353;
int j = x++;
switch (j)
{
case 317: case 353: case 367: System.out.println(Is a prime number.);
case 353: case 363: System.out.println(Is a palindrome.);
break;
default: System.out.println(Invalid value.);
}
}

  1. It is a prime number.

  2. It is a palindrome.

  3. It is a prime number. It is a palindrome.

  4. Invalid value

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

This is the correct choice. The program contains a compilation error. The switch contains two duplicate case values (353) which is not allowed in java. 

Multiple choice

What is the output of the given program code?

public class ScopeTest
{
int z;
public static void main(String[] args)
{
ScopeTest myScope = new ScopeTest();
int z = 6;
System.out.println(myScope.z);
System.out.println(z);
myScope.doStuff();
System.out.println(z);
System.out.println(myScope.z);
}
void doStuff()
{
int z = 5;
doStuff2();
System.out.println(z);
}
void doStuff2()
{
z = 4;
}
}

  1. Garbage value 6 5 6 4

  2. 6 6 5 6 4

  3. 0 6 5 6 4

  4. Compilation error

  5. Run time error
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This is the correct choice. Firstly, myscope.z will be printed, but it has not been initialised to any value. Initially it has a default value 0. So, 0 gets printed. Within main program, z is assigned the value 6. So, z = 6 is printed first. Within doStuff z is assigned 5. DoStuff2 locally sets z to 4 (but MyScope.z is set to 4), but in Dostuff z is still 5. So, z = 5 is printed. Again z is printed within main (with local z set to 6). Then z = 6 is again printed and finally MyScope.z is printed. MyScope.z has been set to 4 within doStuff2(). So, z = 4 gets printed. So, the correct output is 6 5 6 4. This is the correct choice.

Multiple choice

What is the output of the given program?

public class ScopeTest
{
public static void main (String [] args)
{
int[][] xx[][][] = null;
if(xx)
{
System.out.println(xx);
}
else
{
System.out.println("False");
}
}
}

  1. Null

  2. False

  3. Java.Lang.NullPointerException

  4. 0

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

This is the correct choice. Line if(xx) contains an error because we can pass only Boolean value to the if() condition, but here null is passed. So, the compiler will raise an error that “cannot be converted to Boolean”. So, this answer is true.

Multiple choice

What is the output of the given program code?

public class ScopeTest
{
public static void main(String[] args)
{
int [] xx = null;
for(int i=0;i

  1. Null

  2. Java.lang.NullPointerException

  3. 0

  4. Compilation fails

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

This is the correct choice. Java.lang.NullPointerException will be thrown as we are trying to access the array that has been assigned a null value. So, exception will be thrown at run time. So, this is the correct choice.

Multiple choice

How many times is “aa” printed as a part of output?

public class ScopeTest
{
int z;
public static void main(String[] args)
{
String[] table = {"aa", "bb", "cc"};
for (String ss: table)
{
int ii = 0;
while (ii < table.length)
{
System.out.println(ss + ", " + ii);
ii++;
}
}
}
}

  1. Zero times

  2. Once

  3. Twice

  4. Thrice

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

This is the correct choice. The for statement, for (String ss: table), is executed one time for each of the three elements in table. The output comes out to be : aa, 0aa, 1aa, 2bb, 0bb, 1bb, 2cc, 0cc, 1cc, 2. So, “aa” will be printed 3 times, hence this answer is correct.

Multiple choice

What is the output of the given program?

class ScopeTest
{
int x(double d)
{
System.out.println("one");
return 0;
}
String x(double d)
{
System.out.println("two");
return null;
}
double x(double d)
{
System.out.println("three");
return 0.0;
}
public static void main(String[] args)
{
new Overloading().x(4.0);
}
}

  1. One

  2. Two

  3. Three

  4. Compilation fails

  5. Run time error
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This is the correct choice. Overloading of the x method fails as the input argument in all three cases are double. To use overloading of methods, the argument types must be different. Note: The Java programming language supports overloading methods and Java can distinguish between methods with different method signatures. This means that methods within a class can have the same name if they have different parameter lists.