c) is the correct choice because a = 6 - 2 since d = 6 and b = 2. Substitute the numbers for the variables in each answer choice. You will see that all of the answer choices are not equal to 4 except for c).

While the code snippet doesn't return an error, it also doesn't return the answer that you want. Testing string equality requires the method String.equals(). This code snippet uses == to compare two strings. == is a pointer equality function. To get the expected answer (true), the code should return s1.equals(s2).

Answer: II, IV, V

The best way to approach this problem is to divide each conditional into chunks. Then, if there is at least one True in a long list of OR statements, the statement must be true according to Boolean Logic. Likewise, if there is at least one False in a long list of AND's, the statment must be False. For example

False && False && False && True evalulates to False

and

True || False || False ... evaluetes to True

Statement I evalutes to False because False and any number of And's must evaluate to false because False && X evalutes to False, regardless of the value of X.

Statement II evalutes to True because there is one True statement in a list of OR's that make the entire statement True.

Statement III evalutes to False for the same reason Statement I does.

Statement IV is true because although the conditionals in the parentheses are False, the ! (not) operator negates the False to True.

Statement V is true.

In the second condition, it is not a statement of equivalence but a statement of assignment. This would cause it to have a runtime error. If it was to be a == instead of =, the statement would be true.

The main thing to look at for this question is the main loop for the code.  This loop first goes through each of the characters in the String s:

int i = 0; i < s.length(); i++

Next, notice the condition on the inner loop:

vals[i % vals.length]

The modulus on i will yield values that are between 0 and 2 (given the length of vals).  This means that you will loop in the inner loop the following sequence of times:

5,4,2,5,4,2,5,4

This will replicate the given letter at the index (i) in the initial string, using 5,4,2,5, etc as the replication count.  Thus, you will replicate "H" 5 times, "e" 4, etc.  This gives you an output:

HHHHHeeeerrvvvvvaaaaeeuuuuussss

It is easiest to think of the values array as a String: "I love you!!".

Now, the loop is going to run for   or  times.  Notice what it does on each iteration.  It swaps the values at i and values.length - i -1.  Thus, it will do the following swaps:

0 and 11

1 and 10

etc...

This sequence of swaps will eventually reverse the array.  Thus, your output is:

!!uoy evol I

Given that strings cannot be internally modified, you will have to store your result in a new string, namely s2.  Now, for each character in s, you will have to make that charater lower case to begin with:

char c = Character.toLowerCase(s.charAt(i));

Next, for the odd values of i, you will need to make your value to be upper case.  The modulus operator is great for this!  You can use % 2 to find the odd values.  When the remainder of a division by 2 is equal to 1, you know you have an odd value.  Hence, you have the condition:

 if(i % 2 == 1) {...

Then, once you appropriately capitalize, you can place your character on s2.

The easiest way to consider this is by commenting on the correct answer.  You must begin by defining the character array:

char[] vals = {'A','t', ' ', '6',' ','a','m','!'};

Next, you must initialize the string value s to be an empty string.  This is critical.  Otherwise, you can't build your string!

String s = "";

Next, you have the loop.  This goes through the characters and concatenates the values to the variable s. The operation to concatenate the characters is the "+=". This will give you the string value of the array of characters.

The only major issue with this code is the use of the == operator to compare the two strings.  Since the user has input this value, will not have an equality on this test.  You must use the method .equals in order to check whether two strings are equal.  The code should be:

if(books[i].equals(userInput)){

...

(There are some cases in which == will work for string comparison, namely when literals are involved.  However, you should avoid relying on this and always use .equals().)

The compareTo method for strings compares to string objects and returns:

• 0 when they are equal
• Something positive when the given string is alphabetically (really, lexicographically) later than the argument to the method.
• Something negative when the given string is alphabetically (really, lexicographically) before the argument to the method.

Our strings could be put in this order:

"Cursus Philosophicus Thomisticus"

"Happy words for happy people"

"Insane words for an insane world"

"Kritik der reinen Vernunft"

"Logica sive Ars Rationalis"

For each of these, we are asking, "Is it later in order than 'Kittens in a cart'?"  This is true for "Kritik der reinen Vernunft" and "Logica sive Ars Rationalis".  Thus, our output is:

[Logica sive Ars Rationalis, Kritik der reinen Vernunft]

Notice the order—this is due to the order in the original array.