First of all, you can eliminate the following two options very quickly, as the fun method returns a double value:

\[D@4b71bbc9\]

\[4,9,11,19\]

These are trying to trick you into thinking that it returns an array. (The first value is really what would be the output, since there is no toString defined for the double array by default.)

Now, in fun, the loop simply runs through the values in the array, summing those up. Notice, it runs "backwards" through the array, from the end to the beginning. It then divides by the length of the array, giving you the average of the values in the array.

In this code's loop, notice that it takes the absolute value of each element. This is done on the line:

int y = Math.abs(x\[i\]);

This value is then used for the second loop, which goes for y times, each time outputting the value of the given member of the original array—but now with its particular sign value. Thus, even numbers like will be output multiple times (i.e. 8). This is done line by line for each member of the parameter array.

The graphics method takes the 2D array matrix and then runs through each element. This is the point of the double for loop in the method itself. Notice that for each element it does several things. First of all, it is clearly accumulating a value into the variable r. Now, for each element, you are adding:

x\[i\]\[j\] (the current value in the 2D array iteration)

TIMES

(i + 1), or, the current row number. Thus, for the data given you are doing the following:

1* 1 + 1 * 6 + 1 * 7 + 2 * 1 + 2 * 4 + 2 * 5 = 34

Let's look at the main loop in this program:

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

newVals\[i\] = new int\[vals\[i\]\];

for(int j = 0; j < vals\[i\];j++) {

newVals\[i\]\[j\] = vals\[i\] * (j+1);

}

}

The first loop clearly runs through the vals array for the number of items in that array. After this, it creates in the newVals 2D array a second dimension that is as long as the value in the input array vals. Thus, for 41, it creates a 3rd row in newVals that is 41 columns wide.

Then, we go to the second loop. This one iterates from 0 to the value stored in the current location in vals. It places in the given 2D array location the multiple of the value. Think of j + 1. This will go from 1 to 41 for the case of 41 (and likewise for all the values). Thus, you create a 2D array with all the multiples of the initial vals array.

This code is using what are called parallel arrays. The boolean array has one value for each character in the String s. Notice the logic in the loop. There is a condition:

if(!b\[i\]) {

c = Character.toUpperCase(c);

} else {

c = Character.toLowerCase(c);

}

This means that if the boolean is false, then you will make the given letter upper case. (This is because of the ! in the condition. Be careful!) Otherwise, you make it lower case.

Carefully walking through the code, you will get:

i rEAd LoGiC fOr FuN!

When a two dimensional array is created and initialized, the way to access the items inside the matrix is by calling the array with the row and column (i.e. myArray\[ROW\]\[COLUMN\]). Keeping in mind that arrays start at 0, the number four would be in row 1, column 0. Therefore to save that number into the variable "fourth" we'll do the following:

fourth = myArray\[1\]\[0\];

*Note: myArray\[1\]\[0\] is not the same as myArray\[0\]\[1\].

myArray\[1\]\[0\] =4 because it is the item located at row=1 and column = 0.

myArray\[0\]\[1\] =12 because it is the item located at row=0 and column = 1.

Let's look at this line of code:

int * data = new int\[12\]

An int pointer is created and an array of size 12 is assigned to it.

When data\[5\] is called, the "data" is dereferenced and the value of the 6th position is returned.

The only one of the choices that does this is

*(data + 5)

In line above, the pointer is incremented ot the 6th position and is then dereferenced to get the value.

If the code was

*data

The first item at the first position will be returned.

*(data+1)

This will return the item and the second position and so on.

Arrays can be two-dimensional. However, when trying to keep track of keys and values it can become complicated when using an array. HashMaps are the best way to represent data containing keys and values.

In Swift, the variable must be declared first with var then given a name. So now we have var arr then to unwrap, we must add a type var arr: \[Int\] and then initialize. Therefore, we have var arr: \[Int\] = \[\]. var arr = \[\] is technically correct, but the prompt asks you to unwrap the variable.

In Swift, the variable must be declared first with var then given a name. So now we have var arr then to unwrap, we must add a type var arr: \[String\] and then initialize. Therefore, we have var arr: \[String\] = \[\]. var arr = \[\] is technically correct, but the prompt asks you to unwrap the variable.

The code segment will work only when the array has at least one integer. If the array has no negative integers, then the while loop will continue running, incrementing i past the length of the array, in which case an Out of Bounds Exception will occur.

You create a matrix also known as a 2 dimensional array the same way you'd instantiate a normal array except the first array space remains blank and you'd insert the number for the amount of rows. Due to the fact that you want 5 columns and 4 rows, you'd only input the 4 into the second array.

In the second loop, when you remove one of the items, you thus make the ArrayList one element shorter. This means that if you attempt to go through to index 11, you will receive an error at some point, for the list will have "shrunk". This is like an array out of bounds error, though it will be an IndexOutOfBoundsException exception.

In the answers, there are only two issues to consider. On the one hand, look at the loop control statement. It is some variant on:

circle.size()

and

circles.length

For the List types, you need to use the method size(), not length (which you use for arrays). Likewise, one way to extract elements from the ArrayList type is to use the .get method. You must use this and not the brackets \[\] that you use for arrays.

Consider the logic that is in the loop. The if statement will be reached either if you:

1. Have a first character that is H or later in the alphabet (in capital letters).
2. Have a fourth character that is less than d.

The first case applies to "Husserl" and "Hegel." However, the second does not apply to any—not even to "Frederick", for d is equal to d, not less than it!

ArrayLists are lists of objects, and thus cannot store primitive types. If you wish to store primitive types in an ArrayList, container objects such as Double or Integer must instead be used. The object reference type List is a valid reference type for an ArrayList object.

In Java, Array is a fixed length data structure, while ArrayList is a variable length Collection Class (other Collection class members include HashMap and HashSet). This means that an Array cannot change its size once it's made; a whole new Array must be made. ArrayLists can change size at will.

Arrays can reference both primitives (like int) and types (like Integer). ArrayLists can only reference types, not primitives. Through a method called "Autoboxing", it can appear that an ArrayList stores a primitive, but what it really is doing is automatically masking the primitive with the type it's like so that it works with the ArrayList.

During the creation of an Array, the .length value is assigned, so whenever you append .length to the end of an array, you get the length (which will never change for the given Array). For ArrayLists, you use the .size() function, which returns the length of the ArrayList. Because the ArrayList size is variable, it is a function (which is why the parenthesis are there at the end).

Java provides the add() function for ArrayLists, which, as the name implies, adds the argument to the ArrayList. Arrays, on the other hand, do not have functions to add elements. To add something to an Array, you'd call something similar to this:

int[] my_array = new int[10];

my_array[0] = 5;

The correct answer uses a ForEach loop. A ForEach loop is recommended for iterating through Lists because Lists contain iterators. ForEach loops use the iterator to iterate through the List. One of the answers used a regular For loop, while the answer was correct, it was not the best choice.

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.