Compile Time Errors - Computer Science

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Question

What is the error in the following code?

int val1 = -14,val2 = 4;

final int val3 = 9;

double val4 = 4.1;

double val5 = 3.1;

val1 = val2 * val3;

val3 = val1 * 12;

val5 = val1 - val3;

val4 = val2 + val5;

Answer

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;

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Question

Consider the following code:

public static class Rectangle {

private double width, height;

public Rectangle(double w,double h) {

width = w;

height = h;

}

``

public double getArea() {

return width * height;

}

``

public double getPerimeter() {

return 2 * width + 2 * height;

}

}

``

public static class Square extends Rectangle {

public Square(double s) {

super(s,s);

}

}

public static void main(String[] args) {

Rectangle[] rects = new Rectangle[6];

for(int i = 0; i < 6; i++) {

if(i % 2 == 0) {

rects[i] = new Rectangle(i+10,i + 20);

} else {

rects[i] = new Square(i+20);

}

}

Square s = rects[1];

}

What is the error in the code above?

Answer

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]);

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Question

public static void main(String[] args) {

int[] x = {3,4,4,5,17,4,3,1};

int[] y = remove(x,4);

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

System.out.print(y[i] + " ");

}

}

``

public static boolean remove(int[] arr, int val) {

boolean found = false;

int i;

for(i = 0; i < arr.length && !found; i++) {

if(arr[i] == val) {

found = true;

}

}

if(found) {

for(int j = i; j < arr.length;j++) {

arr[j - 1] = arr[j];

}

arr[arr.length - 1] = 0;

}

return found;

}

What is the error in the code above?

Answer

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!

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Question

public interface ServerInstance {

byte[] readBytes();

boolean writeBytes(byte[]b);

boolean wake();

boolean status();

void sleep();

}

``

public class MyHost implements ServerInstance {

boolean running = false;

public boolean wake() {

// Other logic code here...

return running;

}

public boolean status() {

// Other logic code here...

return running;

}

public byte[] readBytes() {

byte[] buffer = null;

// Other logic code here...

return buffer;

}

public void sleep() {

// Other logic code here...

running = false;

}

public byte[] writeBytes(byte[] b) {

// Other logic code here...

return b;

}

// Other methods...

}

What is the error in the code above?

Answer

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.

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Question

public static void foo() {

int x = 10; y = 21, z = 30;

int[] arr = null;

for(int i = 0; i < y; i+= 4) {

arr = new int[i / 5];

}

for(int i = 0; i < x; i++) {

arr[i] = z / i;

}

for(int i = 0; i < z; i++) {

arr[i] = z * i;

}

}

Which of the following lines of code will cause a compile-time error?

Answer

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!

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Question

class Base{};

class Derived : public Base{

public:

void method(){ cout<< "method1
"; }

};

class Derived2 : public Base{

public:

void method() { cout<< "method2
"; }

};

int main(){

Base* bp = new Derived();

Derived2* d2p = bp;

d2p -> method();

}

What is the result of compiling and running the program in C++?

Answer

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.

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Question

Given:

const int x = 10;

Which of the following will compile?

Answer

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

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Question

class Base{

protected:

void method();

};

class Derived : public Base{

};

int main(){

Base b;

b.method(); //Line A

Derived d;

d.method(); //Line B

}

Which of the following is true?

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.

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