If you notice, in our for loop, the integer j is used as the iteration variable. However, no where in the code is that variable defined. To fix this issue this could have been done.

for (int j=0;j<3,j++)

Note the bold here is inserted. We need to define j here. We could have also defined j as a double.

Type ifstream objects are used to read from files, NOT to write to a file. To write to a file you can use ofstream. You can also you fstream to both read and write to and from a file.

line d's error is that system.out.println (i); is not capitalized.

line e's error is that there is a missing semicolon.

If you notice, in our for loop, the integer j is used as the iteration variable. However, no where in the code is that variable defined. To fix this issue this could have been done.

for (int j=0;j<3,j++)

Note the bold here is inserted. We need to define j here. We could have also defined j as a double.

Type ifstream objects are used to read from files, NOT to write to a file. To write to a file you can use ofstream. You can also you fstream to both read and write to and from a file.

line d's error is that system.out.println (i); is not capitalized.

line e's error is that there is a missing semicolon.

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;

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 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 iis 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!

In this problem, Derived1 and Derived2 are children of the Base class. If we take a look at this line:

Base* bp = new Derived();

We are assigning a new Derived class to a base pointer. This will compile. Think of the Base as a larger object because it is the parent, so copying a smaller object into a larger one is acceptable.

Now let's look at this one:

Derived2* d2p = bp;

This line will cause the program to not compile. Since the Base class is considered the "bigger" object, copying a bigger object into a "smaller" one will result in a failure to copy everything over, this is known as a Slicing Problem.

We don't even have to look at the next line because we know that the program wil crash.

First we take a look at the given statement:

const int x = 10;

the const in front of "int" means that x will always hold the value of 10 and it will not change.

Let's observe all the choices.

int *p =&x

This line says to assign the address of x (in memory) to the pointer p. This however, will not compile because int * p is not marked as const. x is marked as a const so this forces int * p to be a const as well.

int * const q = &x

There is a const in this case but it is in the wrong place

const int * r = &x

The const is in the correct place and this is the correct answer

To understand this question, we have to understand what protected method means. A protected method is a method that is accessible to methods inside it's own class as well as it's children. This means that a protected method can be called in the child class.

We can see that method() is called inside main. This should already raise a red flag. A protected class is being called outside of the child class so it will not compile. Even those it's being called on the Base and Derived objects, the calls are not made inside Base and Derived class so neither line will compile.

This loop will indeed run 10 times. However, notice that it begins on 0 and ends on 10. Thus, it will begin by executing:

num *= 0

Which is the same as:

num = num * 0 = 0

Thus, you need to start on 1 for i. However, notice also that you need to go from 1 to 10 inclusive. Therefore, you need to change i < 10 to i <= 10.

Recall that a "mutator" is a method that allows you to change the value of fields in a class. Now, for the setTime, setMinutes, and setSeconds methods, you do not check several cases of errors. For instance, you could give negative values to all of these methods without causing any errors to be noted. Likewise, you could assign a minute or second value that is greater than 60, which could cause errors as well. Finally, you could even make the clock have a value that is greater than "23:59:59", which does not make much sense at all!

The problematic line in the code above is the one that reads:

ret\[i\] = a\[i\];

This is going to work well until the end of the array. The variable i is going to go for the length of the array a. However, the array ret is one less in length. This means that you will overrun your array, getting an ArrayIndexOutOfBoundsException at the very end of the looping. You would need to implement two indices—one of which will not be incremented on that one time when you skip the element that is removed from the array.

The loop is infinite because i never gets incremented to become greater than count.

The loop in this case is,

while (i < count) {

if (i > 0) {

arr\[i\] = count;

}

and the error occurs because there is no terminating factor in the while loop. A terminating factor would be,

while(i<count){

if(i>0, i<10, i++){

arr\[i\] = count;

}

In order to understand the error in this code, you must understand what the loop is doing. It is assigning variable types to the array of Object objects based upon the remainder of dividing the loop control variable _i_ by 4. You thus get a repeating pattern:

Integer, String, Double, String, Integer, String, Double, String, . . .

Now, index 8 will be an Integer (for it has a remainder of 0). This means that when you do the type cast to a String, you will receive a TypeCastException for the line reading:

String s = (String)objects[8];

At this point of the code, it is not possible for a or b to be null. Furthermore, these arrays cannot contain null values, for they are made up of primitive types (int types). These cannot be set to null, as they are not objects. The potential error here is a little bit abstruse, but it is important to note. There is one possible error, namely that the 2D arrays are "ragged"; however we don't need to worry about this for the exam. Still, it is also possible that the array b (presuming that it is not null) is not the same size as the array a. Since we are using a to set our loop counts, this could potentially mean that we overrun the array b by using index values that are too large. (Note, it is also possible that b could be null, causing an error; however, there are no answer choices for that possibility.)