AP Computer Science A : Program Analysis

Study concepts, example questions & explanations for AP Computer Science A

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Example Questions

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Example Question #1 : Unit Testing

The function fun is defined as follows:

public int fun(int[] a)
{
a[a.length - 1] = a[0];
return a[0] + (a[0] % 2);
}
 
What is the value of a[0] after the following code segment is executed?
 
int[] a = {3, 6, 9, 12};
a[0] = fun(a);
Possible Answers:

6

3

12

13

9

Correct answer:

12

Explanation:

The first part of fun assigns the value of the last location of the array to the first location. Then, it returns the a[0] + (a[0] % 2). That last portion will be a "1" if a[0] is odd, and "0" if a[0] is even. Since a[0] == 12, which is even, the expression evaluates to 12 + 0, which is 12.

Example Question #1 : Testing

What are the values of x, y, and z after the following code is executed?
 
int x = 4, y = 3, z;
for (int i = 0; i < 5; i++)
{
     z = x + y;
     y = x - y;
     x = z;
}
Possible Answers:

x == 14, y == 2, z == 14

x == 28, y == 4, z== 4

x == 32, y == 24, z == 32

x == 16, y == 12, z== 16

x == 28, y == 4, z== 28

Correct answer:

x == 28, y == 4, z== 28

Explanation:

The loop will run 5 times. The values of x, y, and z after each run will be as follows:

  • i == 0: x == 7,   y == 1,   z == 7 
  • i == 1: x == 8,   y == 6,   z == 8 
  • i == 2: x == 14, y == 2,   z == 14 
  • i == 3: x == 16, y == 12, z == 16
  • i == 4: x == 28, y == 4,   z == 28

At the end, i will equal 5, and the values will no longer change.

Example Question #1 : Testing

Consider the method
 
public String mystery(String s)
{
String s1 = s.substring(0,1);
String s2 = s.substring(1,2);
String s3 = s.substring(2, s.length() - 2);
String s4 = s.substring(s.length() - 2, s.length() - 1);
String s5 = s.substring(s.length() - 1);
 
     if (s.length() <= 5)
          return s5 + s4 + s3 + s2 + s1;
     else
          return s1 + s2 + mystery(s3) + s4 + s5;
}

What is the output of

System.out.println(mystery("ABNORMALITIES"));

Possible Answers:

SEITILAMRONBA

None of these answers is correct.

SEITRMALIONBA

ABNOILAMRTIES

ABNORMALITIES

Correct answer:

ABNOILAMRTIES

Explanation:

The .substring() method takes the character at the first number in the arguments, and goes through the String until it reaches the second number in the arguments, without copying the character at the second number.

In the first part of mystery(), the Strings s1, s2, s3, s4, and s5 are made and filled. If n = # of characters in s, s1 gets the first character in s, s2 gets the second character in s, s3 gets the third through n-2 characters, s4 gets the n-1 character, and s5 gets the last character.

String s1 = s.substring(0,1);
String s2 = s.substring(1,2);
String s3 = s.substring(2, s.length() - 2);
String s4 = s.substring(s.length() - 2, s.length() - 1);
String s5 = s.substring(s.length() - 1);
 

Let's look at the second portion of mystery(). 

if (s.length() <= 5)
     return s5 + s4 + s3 + s2 + s1;
else
     return s1 + s2 + mystery(s3) + s4 + s5;

The if statement checks the length of s, and if it's less than or equal to 5, it returns a String made from s5, followed by s4, etc. If s were equal to "abcde", then the if would evaluate to true, and would return "edcba". In recursion, this is known as the "base case".

The else statement is for strings that are greater than 5 characters in length. It returns s1, followed by s2, then the result of mystery(s3), then s4 and s5. The fact that it calls itself makes this recursion. 

Let's step through the example. The argument for mystery(), s, is "ABNORMALITIES". After the first part, this is the result:

s1 = "A"
s2 = "B"
s3 = "NORMALITI"
s4 = "E"
s5 = "S"
 
Because s is longer than 5 characters, we take the else, so it returns the following:
 
A + B + mystery(NORMALITI) + E + S
 
Next, we repeat with the new argument. s = NORMALITI, so after the first part, the result is:
 
s1 = "N"
s2 = "O"
s3 = "RMALI"
s4 = "T"
s5 = "I"
 
Because s is longer than 5 characters again, we take the else, so it returns the following:
 
N + O + mystery(RMALI) + T + I
 
Which gets added to the previous return, making it this:
 
A + B + N + O + mystery(RMALI) + T + I + E + S
 
Once again, we repeat with the argument s = RMALI. After the first part, the result is:
 
s1 = "R"
s2 = "M"
s3 = "A"
s4 = "L"
s5 = "I"
 
Because s is less than or equal to 5 characters in length, we take the if this time. It returns the following:
 
I + L + A + M + R
 
We can replace all instances of mystery(RMALI) with the above, so the original return becomes this:
 
A + B + N + O + I + L + A + M + R + T + I + E + S
 
Which gets printed as ABNOILAMRTIES, the answer.
 

Example Question #2 : Testing

Consider the Array

int[] arr = {1, 2, 3, 4, 5};

What are the values in arr after the following code is executed?

for (int i = 0; i < arr.length - 2; i++)
{
int temp = arr[i];
arr[i] = arr[i+1];
arr[i+1] = temp;
}
Possible Answers:

54321

15432

OutOfBoundsException

23451

12345

Correct answer:

23451

Explanation:

We start with an array, arr, of size 5, containing {1, 2, 3, 4, 5}. 

The loop in the code,

for (int i = 0; i < arr.length - 2; i++)

loops through the array up to the second to last cell, given that arr.length - 2 is the index of the second to last cell, and it starts at the first cell. 

Let's look at the code inside the loop.

int temp = arr[i];
arr[i] = arr[i+1];
arr[i+1] = temp;
 
When i = 0,
 
arr[0] == 1
arr[1] == 2
 
temp = 1
arr[0] = 2
arr[1] = 1
 
arr[] == {2, 1, 3, 4, 5}
 
When i = 1
 
arr[1] == 1
arr[2] == 3
 
temp = 1
arr[1] = 3
arr[2] = 1
 
arr[] == {2, 3, 1, 4, 5}

When i = 2

arr[2] == 1
arr[3] == 4
 
temp = 1
arr[2] = 4
arr[3] = 1
 
arr[] == {2, 3, 4, 1, 5}
 
When i = 3
 
arr[3] == 1
arr[4] == 3
 
temp = 1
arr[3] = 3
arr[4] = 1
 
arr[] == {2, 3, 4, 5, 1}

As the loop progresses, it moves whatever value is in arr[0] all the way to the end of the array.

Example Question #1 : Identifying Boundary Cases

Identify a user error that could occur in this program

 

UserInput ui = new UserInput(); // input from the user

int s = (Integer)ui;

System.out.println(s);

Possible Answers:

The user can never do anything wrong

There is nothing wrong

The user input could not be an integer

The parseInt statement is incorrect

Correct answer:

The user input could not be an integer

Explanation:

The user could input a string and the cast to an (Integer) would cause a runtime exception. If there is a runtime exception, the program will stop and open up vulnerabilities to hackers. Once a hacker knows how to halt a program, they can start input bad data to see a database schema to collect data. One way to fix this would be using a utility method such as parseInt(ui). 

Example Question #1 : Program Analysis

Consider the following code:

public static class Rectangle {

     private double width, height;

     public Rectangle(double w,double h) {

          width = w;

          height = h;

     }

    

     public double getArea() {

          return width * height;

     }

    

     public double getPerimeter() {

          return 2 * width + 2 * height;

     }

}

 

public static class Square extends Rectangle {

     public Square(double s) {

          super(s,s);

     }

}

public static void main(String[] args) {

     Rectangle[] rects = new Rectangle[6];

     for(int i = 0; i < 6; i++) {

          if(i % 2 == 0) {

               rects[i] = new Rectangle(i+10,i + 20);

          } else {

               rects[i] = new Square(i+20);

          }

     }

     Square s = rects[1];

}

What is the error in the code above?

Possible Answers:

You need to use "implements", not "extends" for the class Square.

There is an error in the declaration of the Rectangle array.

There is an array overrun.

The final assignment operation cannot be done.

You cannot assign a Square object to an array of Rectangle objects.

Correct answer:

The final assignment operation cannot be done.

Explanation:

This code fills up the 6 member array with alternating Rectangle and Square objects. You can do this because the Square class is a subclass of Rectangle. That is, since Squares are Rectangles, you can store Square objects in Rectangle variables. However, even though rects[1] is a square, you CANNOT immediately reassign that to a Square object. The code has now come to consider all of the objects in the array as being Rectangle objects. You would need to explicitly type cast this to get the line to work:

Square s = (Square)(rects[1]);

Example Question #1 : Program Analysis

What is the error in the following code?

int val1 = -14,val2 = 4;

final int val3 = 9;

double val4 = 4.1;

double val5 = 3.1;

val1 = val2 * val3;

val3 = val1 * 12;

val5 = val1 - val3;

val4 = val2 + val5;

Possible Answers:

You cannot perform mixed addition or subtraction with doubles and integers.

You must define val2 on its own line.

You cannot assign the new integer value to val3.

You cannot assign the result of an integer subtraction to val5.

You cannot multiply val3 by val2.

Correct answer:

You cannot assign the new integer value to val3.

Explanation:

The only error among the options given is the fact that this code assigns a new value to the variable val3, which is defined as a constant. (This is indicated by the keyword final before the rest of its declaration.) You cannot alter constants once they have been declared. Thus, the following line will cause a compile-time error:

val3 = val1 * 12;

Example Question #1 : Debugging

public static void main(String[] args) {

       int[] x = {3,4,4,5,17,4,3,1};

       int[] y = remove(x,4);

       for(int i = 0; i < y.length; i++) {

              System.out.print(y[i] + " ");

       }

}

 

public static boolean remove(int[] arr, int val) {

       boolean found = false;

       int i;

       for(i = 0; i < arr.length && !found; i++) {

              if(arr[i] == val) {

                     found = true;

              }

       }

       if(found) {

              for(int j = i; j < arr.length;j++) {

                     arr[j - 1] = arr[j];

              }

              arr[arr.length - 1] = 0;

       }

       return found;

}

What is the error in the code above?

Possible Answers:

There is a type mismatch in one of the assignments

There is no error in the code

There are multiple options to delete, which is not supported by the function

In the remove method, there is a variable that has not been initialized before being used

The output loop in main will overrun the array y

Correct answer:

There is a type mismatch in one of the assignments

Explanation:

The problematic line is this one:

int[] y = remove(x,4);

Notice that the variable y is defined as an array. Now, it is tempting to think (without looking too closely) that the remove method returns the array after the removal has been accomplished; however, this is not how the logic works in the remove method. Instead, it returns a boolean indicating whether or not this removal was successful or not (i.e. it tells you whether or not it actually found the value). Therefore, you cannot make an assignment like the one above, for the two types are not the same. That is, y is an integer array, while remove returns a boolean value!

Example Question #241 : Computer Science

public interface ServerInstance {

            byte[] readBytes();

            boolean writeBytes(byte[]b);

            boolean wake();

            boolean status();

            void sleep();

}

 

public class MyHost implements ServerInstance {

            boolean running = false;

            public boolean wake() {

                        // Other logic code here...

                        return running;

            }

            public boolean status() {

                        // Other logic code here...

                        return running;

            }

            public byte[] readBytes() {

                        byte[] buffer = null;

                        // Other logic code here...

                        return buffer;

            }

            public void sleep() {

                        // Other logic code here...

                        running = false;

            }

            public byte[] writeBytes(byte[] b) {

                        // Other logic code here...

                        return b;

            }

            // Other methods...

}

What is the error in the code above?

Possible Answers:

The readBytes method initializes the variable buffer to null

There is no error

The writeBytes method is not properly defined

The word extends should be used, not implements

The sleep method does not return anything

Correct answer:

The writeBytes method is not properly defined

Explanation:

When you implement an interface, all of the methods defined in that interface must be written in the class that is proposing to be such an implementation. (Or, if they are not implemented there, you need to involve abstract classes—but that is not our concern here.) The methods must match the prototypes proposed in the interface. In the example code, ServerInstance has a method writeBytes that returns a boolean value. However, the MyHost class has implemented this method as returning a byte[] value.  Since you cannot have different types of return values for methods with the same parameter set, Java interprets this as being the proposed implementation for writeBytes(byte[] b), and this method must return a boolean if MyHost is to implement ServerInstance.

Example Question #1 : Compile Time Errors

public static void foo() {

            int x = 10; y = 21, z = 30;

            int[] arr = null;

            for(int i = 0; i < y; i+= 4) {

                        arr = new int[i / 5];

            }

            for(int i = 0; i < x; i++) {

                        arr[i] = z / i;

            }

            for(int i = 0; i < z; i++) {

                        arr[i] = z * i;

            }

}

Which of the following lines of code will cause a compile-time error?

Possible Answers:

arr = new int[i / 5];

arr[i] = z * i;

arr[i] = z / i;

None of these lines will cause a compile-time error

int x = 10; y = 21, z = 30;

Correct answer:

int x = 10; y = 21, z = 30;

Explanation:

An error in compilation occurs before any code even attempts to execute. Thus, it is primarily a syntactical error in the code. In the selection above, the line

int x = 10; y = 21, z = 30;

has a semicolon right after 10. This causes an error in the code directly following on this, for the code

y = 21, z = 30;

is not valid Java code.

There are other errors in this code. arr[i] = z / i; will cause a divide by 0 error when i is 0. Also, arr[i] = z * i; will overrun the bounds of the array arr. However, these are not compile-time errors—i.e. errors that occur before the code is even able to run!

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