1.1.6. Pretest for the AP CSA Exam¶

Please try the following pretest. We don’t expect you to know the answers to these questions at all yet! So don’t worry about it if you don’t know the answers. If you have no idea on the answer, it is okay to skip questions or to make your best guess.

You will take the same test at the end of the class, and the goal is to see how much you improve. We hope that everyone will see a big improvement!

Please make sure you are logged into Runestone before beginning. Click the button when you are ready to begin the exam. You can only take the exam once. There are no time limits, but it will keep track of how much time you take. Click on the button to go to the next question. Click on the button to go to the previous question. Use the number buttons to jump to a particular question. Click the button to pause the exam (you will not be able to see the questions when the exam is paused). Click on the button after you have answered all the questions. The number correct, number wrong, and number skipped will be displayed.

1-1-6-1: Given the following code segment, what is printed when it is executed?

String test = "123456";
for (int index = 0; index < test.length() - 1; index = index + 2)
{
    System.out.print(test.substring(index,index+2));
}

112233445566
This would be true if the loop was printing each character twice and was incrementing the index by 1, but it prints two characters at a time and increments the index by 2.
123456
This will loop through the string and print two characters at a time. The first time through the loop index = 0 and it will print "12". The second time through the loop index = 2 and it will print "34". The third time through the loop index = 4 and it will print "56". Remember that the substring method that takes two integer values will start the substring at the first value and include up to the character before the second value.
123234345456
This would be true if the loop was loop printing three characters at a time, but it prints two characters at a time.
1223344556
This would be true if the index was incrementing by 1 instead of 2.
Nothing will be printed due to an IndexOutOfBoundsException.
This would be true if the loop stopped when index was less than the string length instead of one less than the string length.

1-1-6-2: Consider the following data field and method findLongest. Method findLongest is intended to find the longest consecutive block of the value target occurring in the array nums; however, findLongest does not work as intended. For example, if the array nums contains the values [7, 10, 10, 15, 15, 15, 15, 10, 10, 10, 15, 10, 10], the call findLongest(10) should return 3, the length of the longest consecutive block of 10s. Which of the following best describes the value returned by a call to findLongest?

private int[] nums;
public int findLongest(int target)
{
   int lenCount = 0;
   int maxLen = 0;

   for (int k = 0; k < nums.length; k++)
   {
      if (nums[k] == target)
      {
         lenCount++;
      }

      else
      {
         if (lenCount > maxLen)
         {
            maxLen = lenCount;
         }
      }
   }

   if (lenCount > maxLen)
   {
      maxLen = lenCount;
   }

   return maxLen;
}

It is the length of the array nums.
This can not be true. There is no nums.length in the code and the only count happens lenCount is incremented when nums[k] == target.
It is the length of the first consecutive block of the value target in nums.
It does not reset the count ever so it just counts all the times the target value appears in the array.
It is the number of occurrences of the value target in nums.
The variable lenCount is incremented each time the current array element is the same value as the target. It is never reset so it counts the number of occurrences of the value target in nums. The method returns maxLen which is set to lenCount after the loop finishes if lenCount is greater than maxLen.
It is the length of the shortest consecutive block of the value target in nums.
It does not reset the count ever so it just counts all the times the target value appears in the array.
It is the length of the last consecutive block of the value target in nums.
It does not reset the count ever so it just counts all the times the target value appears in the array.

1-1-6-3: Given the following code segment, what are the values of var1 and var2 after the while loop finishes?

int var1 = 0;
int var2 = 2;

while ((var2 != 0) && ((var1 / var2) >= 0))
{
   var1 = var1 + 1;
   var2 = var2 - 1;
}

var1=1, var2=1
This would be true if the body of the while loop only executed one time, but it executes twice.
var1=3, var2=-1
This would be true if the body of the while loop executed 3 times, but it exectues twice.
var1=0, var2=2
This would be true if the body of the while loop never executed. This would have happened if the while check was if var1 != 0 instead of var2 != 0.
var1=2, var2=0
The loop starts with var1=0 and var2=2. The while checks that var2 isn't 0 (2!=0) and that var1 / var2 is greater than or equal to zero (0/2=0) so this is equal to zero and the body of the while loop will execute. The variable var1 has 1 added to it for a new value of 1. The variable var2 has 1 subtracted from it for a value of 1. At this point var1=1 and var2=1. The while condition is checked again. Since var2 isn't 0 (1!=0) and var1/var2 (1/1=1) is >= 0 so the body of the loop will execute again. The variable var1 has 1 added to it for a new value of 2. The variable var2 has 1 subtracted from it for a value of 0. At this point var1=2 and var2=0. The while condition is checked again. Since var2 is zero the while loop stops and the value of var1 is 2 and var2 is 0.
The loop won't finish executing because of a division by zero.
The operation 0 / 2 won't cause a division by zero. The result is just zero.

1-1-6-4: At a certain high school students receive letter grades based on the following scale: 93 or above is an A, 84 to 92 inclusive is a B, 75 to 83 inclusive is a C, and below 75 is an F. Which of the following code segments will assign the correct string to grade for a given integer score?

I.  if (score >= 93)
       grade = "A";
    if (score >= 84 && score <= 92)
       grade = "B";
    if (score >= 75 && score <= 83)
       grade = "C";
    if (score < 75)
       grade = "F";

II. if (score >= 93)
       grade = "A";
    if (score >= 84)
       grade = "B";
    if (score >= 75)
       grade = "C";
    if (score < 75)
       grade = "F";

III. if (score >= 93)
        grade = "A";
     else if (score >= 84)
        grade = "B";
     else if (score >= 75)
        grade = "C";
     else
        grade = "F";

I and III only
Choice I uses multiple if's with logical ands in the conditions to check that the numbers are in range. Choice Choice II won't work since if you had a score of 94 it would first assign the grade to an "A" but then it would execute the next if and change the grade to a "B" and so on until the grade was set to a "C". Choice III uses ifs with else if to make sure that only one conditional is executed.
II only
Choice II won't work since if you had a score of 94 it would first assign the grade to an "A" but then it would execute the next if and change the grade to a "B" and so on until the grade was set to a "C". This could have been fixed by using else if instead of just if.
III only
Choice III is one of the correct answers. However, choice I is also correct. Choice I uses multiple if's with logical ands in the conditions to check that the numbers are in range. Choice III uses ifs with else if to make sure that only one conditional is executed.
I and II only
Choice II won't work since if you had a score of 94 it would first assign the grade to an "A" but then it would execute the next if and change the grade to a "B" and so on until the grade was set to a "C". This could have been fixed by using else if instead of just if.
I, II, and III
Choice II won't work since if you had a score of 94 it would first assign the grade to an "A" but then it would execute the next if and change the grade to a "B" and so on until the grade was set to a "C". This could have been fixed by using else if instead of just if.

1-1-6-5: Given the following code segment, which of the following is this equivalent to?

if ( x > 0) x = -x;
if (x < 0) x = 0;

x = 0;
No matter what x is set to originally, the code will reset it to 0.
if (x > 0) x = 0;
Even if x is < 0, the above code will set it to 0.
if (x < 0) x = 0;
Even if x is > than 0 originally, it will be set to 0 after the code executes.
if (x > 0) x = -x; else x = 0;
The first if statment will always cause the second to be executed unless x already equals 0, such that x will never equal -x
if ( x < 0) x = 0; else x = -1;
The first if statement will always cause the second to be executed unless x already equals 0, such that x will never equal -x

1-1-6-6: Susan is 5 years older than Matt. Three years from now Susan’s age will be twice Matt’s age. What should be in place of condition in the code segment below to solve this problem?

for (int s = 1; s <= 100; s++) {
   for (int m = 1; m <= 100; m++) {
      if (condition)
         System.out.println("Susan is " + s + " and Matt is " + m);
   }
}

(s == m - 5) && (s - 3 == 2 * (m - 3))
This would be true if Susan was 5 years younger than Matt and three years ago she was twice his age. But, how could she be younger than him now and twice his age three years ago?
(s == (m + 5)) && ((s + 3) == (2 * m + 3))
This is almost right. It has Susan as 5 years older than Matt now. But the second part is wrong. Multiplication will be done before addition so (2 * m + 3) won't be correct for in 3 years Susan will be twice as old as Matt. It should be (2 * (m + 3)) or (2 * m + 6).
s == (m - 5) && (2 * s + 3) == (m + 3)
This can't be right because Susan is 5 years older than Matt, so the first part is wrong. It has susan equal to Matt's age minus 5 which would have Matt older than Susan.
s == m + 5 && s + 3 == 2 * m + 6
Susan is 5 years older than Matt so s == m + 5 should be true and in 3 years she will be twice as old so s + 3 = 2 * (m + 3) = 2 * m + 6.
None of the answers are correct
The answer is s == m + 5 && s + 3 == 2 * m + 6.

1-1-6-7: Given the following code segment, what is printed when it executes?

 public static void test()
 {
    int num = 0;
    while(num <= 14)
    {

       if(num % 3 == 1)
       {
          System.out.print("1 ");
       }

       else if (num % 3 == 2)
       {
          System.out.print("2 ");
       }

       else
       {
          System.out.print("0 ");
       }

       num += 2;
    }
}

0 1 2 0 1 2 0 1
The second time through the loop the value of num is 2 and 2 % 3 is 2 not 1.
0 2 1 0 2 1 0 2
The while loop will iterate 8 times. The value of num each time through the loop is: 0, 2, 4, 6, 8, 10, 12, and 14. The corresponding remainder operator of 3 is: 0, 2, 1, 0, 2, 1, 0, 2, which is print to the console.
0 2 1 0 2 1 0 2 1
The loop will iterate 8 times not 9. When the value of num exceeds 14, num will no longer be evaluated against the conditional statements. The remainder operator of 3 will be evaluated on the num values of 0, 2, 4, 6, 8, 10, 12 and 14.
2 1 0 2 1 0 2 1
The value of num the first time through the loop is 0 so the first remainder is 0 not 2. This would be true if the value of num was 2 to start.
0 2 1 0 2 1 0
This would be true if the loop stopped when the value of num was less than 14 but it is less than or equal to 14.

1-1-6-8: Given the following incomplete class declaration, which of the following can be used to replace the missing code in the advance method so that it will correctly update the time?

public class TimeRecord
{
    private int hours;
    private int minutes; // 0<=minutes<60

    public TimeRecord(int h, int m)
    {
        hours = h;
        minutes = m;
    }

    // postcondition: returns the
    // number of hours
    public int getHours()
    {
        /* implementation not shown */
    }

    // postcondition: returns the number
    // of minutes; 0 <= minutes < 60
    public int getMinutes()
    {
        /* implementation not shown */
    }

    // precondition: h >= 0; m >= 0
    // postcondition: adds h hours and
    // m minutes to this TimeRecord
    public void advance(int h, int m)
    {
        hours = hours + h;
        minutes = minutes + m;
        /* missing code */
    }

    // ... other methods not shown

}

hours = hours + minutes / 60; minutes = minutes % 60;
This will update the hours and minutes correctly. It will add the floor of the division of minutes by 60 to hours and then set minutes to the remainder of the division of minutes by 60.
minutes = minutes % 60;
This won't add to hour so it can't be correct. It will set minutes to the remainder of dividing minutes by 60 so minutes will be set correctly.
minutes = minutes + hours % 60;
This will set the minutes to the minutes plus the remainder of dividing the hours by 60.
hours = hours + minutes % 60; minutes = minutes / 60;
This will set hours to hours plus the remainder of dividing minutes by 60 and then set minutes to the number of hours (int division of minutes by 60).
hours = hours + minutes / 60;
This will correctly update the hours, but not update the minutes.

1-1-6-9: Which of the following expressions is equivalent to the following?

!(c || d)

(c || d)
NOTing an OR expression does not result in the same values ORed.
(c && d)
You do negate the OR to AND, but you also need to negate the values of d and d.
(!c) && (!d)
NOTing (negating) an OR expression is the same as the AND of the individual values NOTed (negated). See De Morgans laws.
!(c && d)
This would be equivalent to (!c || !d)
(!c) || (!d)
This would be equivalent to (!(c && d))

1-1-6-10: Which of the following will cause an infinite loop when temp is greater than zero and a is an array of integers.

for (int k = 0; k < a.length; k++ )
{
   while (a[k] < temp)
   {
      a[k] *= 2;
   }
}

The values don't matter this will always cause an infinite loop.
An infinite loop will not always occur in this program segment. It occurs when at least one value in a is less than or equal to 0.
Whenever a has values larger than temp.
Values larger then temp will not cause an infinite loop.
When all values in a are larger than temp.
Values larger then temp will not cause an infinite loop.
Whenever a includes a value that is equal to zero.
When a contains a value that is equal to zero then multiplying that value by 2 will always be 0 and will never make the result larger than the temp value (which was set to some value > 0), so an infinite loop will occur.
Whenever a includes a value equal to temp.
Values equal to temp will not cause the infinite loop.

1-1-6-11: Given the following method declaration, and int[] a = {8, 3, 1}, what is the value in a[1] after m1(a); is run?

public static int m1(int[] a)
{
   a[1]--;
   return (a[1] * 2);
}

4
This would be true if it was return (a[1] *= 2);
2
The statement a[1]--; is the same as a[1] = a[1] - 1; so this will change to 3 to 2. The return (a[1] * 2) does not change the value at a[1].
16
This would be true if it was return (a[0] *= 2);
7
This would be true if it was a[0]--;
3
This can't be true because a[1]--; means the same as a[1] = a[1] - 1; so the 3 changes to 2. Parameters are all pass by value in Java which means that a copy of the value is passed to a method. But, since an array is an object a copy of the value is a copy of the reference to the object. So changes to objects in methods are permanent.

1-1-6-12: Given the following code segment, what will the value of s1 be after this executes?

String s1 = "Hi There";
String s2 = s1;
String s3 = s2;
String s4 = s1;
s2 = s2.toLowerCase();
s3 = s3.toUpperCase();
s4 = null;

Hi There
Strings are immutable meaning that any changes to a string creates and returns a new string, so the string referred to by s1 does not change
hi there
This would only be correct if we had s1 = s2; after s2.toLowerCase(); was executed. Strings are immutable and so any change to a string returns a new string.
HI THERE
This would be correct if we had s1 = s3; after s3.toUpperCase(); was executed. Strings are immutable and so any change to a string returns a new string.
null
This would be true if we had s1 = s4; after s4 = null; was executed. Strings are immutable and so any changes to a string returns a new string.
hI tHERE
Strings are immutable and so any changes to a string returns a new string.

1-1-6-13: Which of the following is printed as the result of the call mystery(1234);?

//precondition:  x >=0
public void mystery (int x)
{

   System.out.print(x % 10);

   if ((x / 10) != 0)
   {
      mystery(x / 10);
   }

   System.out.print(x % 10);
}

Many digits are printed due to infinite recursion.
When the recursive call to mystery(1) occurs (the 4th call to mystery), the division of x /10 equals .01--this becomes 0 because this is integer division and the remainder is thrown away. Therefore the current call will be completed and all of the previous calls to mystery will be completed.
3443
The first call to mystery with the integer 1234 will print 1234 % 10. The '%' means remainder. The remainder of 1234 divided by 10 is 4 so the first thing printed must be 4.
12344321
The first call to mystery with the integer 1234 will print 1234 % 10. The '%' means remainder. The remainder of 1234 divided by 10 is 4 so the first thing printed must be 4.
1441
The first call to mystery with the integer 1234 will print 1234 % 10. The '%' means remainder. The remainder of 1234 divided by 10 is 4 so the first thing printed must be 4.
43211234
This has a recursive call which means that the method calls itself when (x / 10) is greater than or equal to zero. Each time the method is called it prints the remainder of the passed value divided by 10 and then calls the method again with the result of the integer division of the passed number by 10 (which throws away the decimal part). After the recursion stops by (x / 10) == 0 the method will print the remainder of the passed value divided by 10 again.

1-1-6-14: Under which of these conditions will a sequential search be faster than a binary search?

The search value is not in the array
If the search value is not in the array, a sequential search will have to check every item in the array before failing, a binary search will be faster.
The search value is the last element in the array
In this case a sequential search will have to check every element before finding the correct one, whereas a binary search will not.
The value is in the middle of the array.
Results will differ depending on the exact location of the element, but Binary Search will still find the element faster while Sequential will have to check more elements.
The search value is the first element in the array.
Only when the search value is the first item in the array, and thus the first value encountered in sequential search, will sequential be faster than binary.
Sequential Search can never be faster than Binary Search.
When the search value is the first element, Sequential will always be faster, as it will only need to check one element.

1-1-6-15: Given the following code segment, what will be printed when it is executed?

List<Integer> list1 = new ArrayList<Integer>();
list1.add(new Integer(1));
list1.add(new Integer(2));
list1.add(new Integer(3));
list1.set(2, new Integer(4));
list1.add(2, new Integer(5));
list1.add(new Integer(6));
System.out.println(list1);

[1, 2, 3, 4, 5]
The set replaces the 3 with the 4 so this can't be right
[1, 2, 4, 5, 6]
The add with an index of 2 and a value of 5 adds the 5 at index 2 not 3. Remember that the first index is 0.
[1, 2, 5, 4, 6]
The add method that takes just a value as a parameter adds that value to the end of the list. The set replaces the value at that index with the new value. The add with parameters of an index and a value puts the passed value at that index and moves any existing values by one index to the right (increments the index). So the list looks like: 1 // add 1 1 2 // add 2 1 2 3 // add 3 1 2 4 // set index 2 to 4 1 2 5 4 // add 5 to index 2 (move rest right) 1 2 5 4 6 // add 6 to end
[1, 5, 2, 4, 6]
The add with an index of 2 and a value of 5 adds the 5 at index 2 not 1. Remember that the first index is 0.
[1, 6, 2, 4, 5]
How did the 6 get in position 2?

1-1-6-16: Given the following code segment, What are the contents of mat after the code segment has been executed?

int [][] mat = new int [3][4];
for (int row = 0; row < mat.length; row++)
{

   for (int col = 0; col < mat[0].length; col++)
   {
      if (row < col)
         mat[row][col] = 1;
      else if (row == col)
         mat[row][col] = 2;
      else
         mat[row][col] = 3;
  }
}

{ {2 1 1 1}, {3 2 1 1}, {3 3 2 1}}
When you create a 2-d array the first value is the number of rows and the second is the number of columns. This code will put a 1 in the array when the row index is less than the column index and a 2 in the array when the row and column index are the same, and a 3 in the array when the row index is greater than the column index.
{ {2 3 3}, {1 2 3}, {1 1 2}, {1 1 1}}
This would be true if the first value when you create a 2-d array was the number of columns and the second was the number of rows. Also you would need to set the value to 3 when the column index was greater than the row and a 1 when the row index was greater than the column index.
{ {2 1 1}, {3 2 1}, {3 3 2}, {3 3 3}}
This would be true if the first value when you create a 2-d array was the number of columns and the second was the number of rows.
{ {2 3 3 3}, {1 2 3 3}, {1 1 2 3}}
This would be true if you set the value to 3 when the column index was greater than the row and a 1 when the row index was greater than the column index.
{ {1 1 1 1}, {2 2 2 2}, {3 3 3 3}}
This would be true if you set the value to the row index.

1-1-6-17: Given the following class declarations, and assuming that the following declaration appears in a client program: Base b = new Derived();, what is the result of the call b.methodOne();?

public class Base
{

    public void methodOne()
    {
        System.out.print("A");
        methodTwo();
    }

    public void methodTwo()
    {
        System.out.print("B");
    }
}

public class Derived extends Base
{

    public void methodOne()
    {
        super.methodOne();
        System.out.print("C");
    }

    public void methodTwo()
    {
        super.methodTwo();
        System.out.print("D");
    }
}

AB
This would be true if the object was created of type Base using new Base. But the object is really a Derived object. So all methods are looked for starting with the Derived class.
ABDC
Even though b is declared as type Base it is created as an object of the Derived class, so all methods to it will be resolved starting with the Derived class. So the methodOne() in Derived will be called. This method first calls super.methodOne so this will invoke the method in the superclass (which is Base). So next the methodOne in Base will execute. This prints the letter "A" and invokes this.methodTwo(). Since b is really a Derived object, we check there first to see if it has a methodTwo. It does, so execution continues in Derived's methodTwo. This method invokes super.methodTwo. So this will invoke the method in the super class (Base) named methodTwo. This method prints the letter "B" and then returns. Next the execution returns from the call to the super.methodTwo and prints the letter "D". We return to the Base class methodOne and return from that to the Derived class methodOne and print the letter "C".
ABCD
After the call to methodOne in the super class printing "A", the code continues with the implicit this.methodTwo which resolves from the current object's class which is Derived. methodTwo in the Derived class is executed which then calls super.methodTwo which invokes printin "B" from methodTwo in the Base class. Then the "D" in the Derive methodTwo is printed. Finally the program returns to methodOne in the Derived class are prints "C".
ABC
The call to methodTwo in super.methodOne is to this.methodTwo which is the method from the Derived class. Consequently the "D" is also printed.
Nothing is printed due to infinite recursion.
This is not an example of recursion. No method is called from within itself.

1-1-6-18: Given the following code segment, what are the values of a and b after the for loop finishes?

int a = 10, b = 3, t;
for (int i=1; i<=6; i++)
{
   t = a;
   a = i + b;
   b = t - i;
}

a = 6 and b = 7
This would be true if the loop stopped when i was equal to 6.
a = 6 and b = 13
Actually i = 6 and t = 6 and a = 13 after the loop finishes.
a = 6 and b = 0
Actually i = 6 and t = 6 and b = 0 after the loop finishes.
a = 0 and b = 13
Actually a = 13 and b = 0 after the loop finishes.
a = 13 and b = 0
The variable i loops from 1 to 6 and each time the values are as follows: i = 1, t = 10, a = 4, b = 9, i = 2, t = 4, a = 11, b =2, i = 3, t = 11, a = 5, b = 8, i = 4, t = 5, a = 12, b = 1, i = 5, t = 12, a = 6, b = 7, i = 6, t = 6, a = 13, b = 0

1-1-6-19: What is data encapsulation and how does Java implement it?

Data (fields) can be directly accessed by all code in all classes.
Encapsulation is making data private so only code in the same class has direct access.
Data (fields) can be hidden inside of an object using the abstract visibility modifier.
There is no abstract visibility modifier. You cannot use the keyword abstract on variable declarations.
Data (fields) can be hidden inside an object using the visibility modifier private.
This is the definition of encapsulation and this is done in Java using private (a member is directly accessible only in the class that defines it) and protected (a member is directly accessible only within code in the same package and in subclasses).
Data (fields) are directly accessible by objects in the same package and in subclasses.
Encapsulation means that only code in the defining class has direct access. The visibility modifier protected gives diredct access to code in classes in the same package and subclasses.
Data (fields) are directly accessible by objects in the same package.
Encapsulation means that only code in the defining class has direct access. The default package access gives direct access to code in classes in the same package.

1-1-6-20: Which of the following reasons for using an inheritance heirarchy are valid?

I.   Methods from a superclass can be used in a subclass without
     rewriting or copying code.
II.  Objects from subclasses can be passed as arguments to a method
     designed for the superclass
III. Objects from subclasses can be stored in the same array
IV.  All of the above
V.   None of the above

V.
In fact, all of the reasons listed are valid. Subclasses can reuse methods written for superclasses without code replication, subclasses can be stored in the same array, and passed as arguments to methods meant for the superclass. All of which make writing code more streamlined.
I and II
III is also valid. In some cases you might want to store subclasses together in a single array, and inheritance allows for this.
I and III
II is also valid. In some cases a single method is applicable for a number of subclasses, and inheritance allows you to pass objects of the subclasses to the same method instead of writing individual methods for each subclass.
IV
All of these are valid reasons to use an inheritance heirarchy.
I only
II and III are also valid, in some cases a single method is applicable for a number of subclasses, and inheritance allows you to pass all the subclasses to the same method instead of writing individual methods for each subclass and you might want to store subclasses together in a single array, and inheritance allows for this.

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1.1.6. Pretest for the AP CSA Exam¶