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.

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.

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.

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:

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.

When i = 2

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

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). 

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]);

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;

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!

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.

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!