So this is basically my code
abstract class B
{
int x = 3;
B()
{
x+=2;
System.out.print("-x" + x + "\n"); // print -x5
x++; // 5 then 6
}
abstract int calculate();
abstract int calculate(int i);
}
class A extends B
{
int x = 2;
A()
{
System.out.print("-x" + calculate(2)+"\n");
}
#Override
int calculate()
{
return x;
}
#Override
int calculate(int i)
{
return(calculate()+i);
}
}
public class Test2 extends A
{
Test2()
{
x+=3;
}
#Override
int calculate()
{
return x + 6;
}
public static void main(String[] args) {
Test2 sub = new Test2();
System.out.print("-x" + sub.calculate()+"\n");
}
}
My problem here is after digging up about variable hiding I learned that if a instance variable is of the same name in both parent class and child class then the childclass hides the instance variable of the parent class. Also I have the prior knowledge that variables cannot be overridden when child class inherits parent class.
So now coming to the problem, in the code when A extends to B, why does the print statement inside constructor A() gives a value -x10? shouldn't it be -x8?? I dont understand how the variable is being changed here. I am new to java so any kind of knowledge will be greatly appreciated. :)
Ok so I have done some debugging and found that the calculate(void) method in class A returns 8. But how is that possible shouldn't it return 6? Please help!
The reason it prints -x10 is because A::calculate(2) calls the Test2::calculate(), which uses A::x to do the calculation.
The sequence of calls that happens is the following:
Test2() {
A()
B() {
B::x = 3
B::x += 2
System.out.print("-x" + x + "\n"); // print -x5
B::x++ // B::x is now 6
}
A::x = 2
System.out.print("-x" + calculate(2)+"\n")
A::calculate(2) {
return(calculate()+2);
Test2::calculate() {
return A::x + 6; // A::x is 2 here, so returns 8
}
} // returns calculate()+2, so returns 10
}
A::x += 3
}
I hope this is just code to test things out, because you should never allow this to happen in real code. You should never allow a method of a subclass to be called from the constructor of a base class, because the subclass is not initialised at that time. The Java compiler does its best to prevent that, but sometimes it does not detect it.
It is returning 8 because:
The line you called System.out.print("-x" + sub.calculate()+"\n"); in class A calls
#Override
int calculate()
{
return x + 6;
}
in class A still, which is incrementing the instance variable int x = 2 in class A . this variable overwrote the one in class B
Hence 2+6 = 8
Related
The equals() method should check if the dimensions of the first box and the cube are the same. How to fix it? It currently does not work.
The program returns the message "illegal start of type" at if. I am new to this plz help
public class testNew
{
public static void main (String []args)
{
Rectangle3 one = new Rectangle3(5,20);
Box3 two = new Box3(4,4,4);
Box3 three = new Box3(4,10,5);
Cube3 four = new Cube3(4,4,4);
showEffectBoth(one);
showEffectBoth(two);
showEffectBoth(three);
showEffectBoth(four);
}
public static String showEffectBoth(Rectangle3 r)
{
return System.out.println(r);
}
boolean b = two.equals(four);
if (b == true)
{
System.out.println("Box and cube have the same dimensions");
}
}
public class Rectangle3
{
// instance variables
int length;
int width;
public Rectangle3(int l, int w)
{
length = l;
width = w;
}
public int getLength()
{
return length;
}
public int getWidth()
{
return width;
}
public String toString()
{
return getClass().getName() + " - " + length + " X " + width;
}
public boolean equals(Rectangle3 obj)
{
if ((getLength().equals(obj.getLength()) && getWidth().equals(obj.getWidth())))
return true;
else
return false;
}
}
First, regarding the compiler error you have, it has nothing to do with the equals() method. It's only because all of the code below, should be inside your main method as it's the only part where you are declaring the variablestwo and four:
boolean b = two.equals(four);
if (b == true) {
System.out.println("Box and cube have the same dimensions");
}
Notice also, that the Rectangle3 class shouldn't be in the same file as testNew as both are declared public, if you want to use both of them in the same file then you need to remove the public declration from one of them (the one you will not use as filename)
Second, your equals() method is technically correct (I guess functionally as well) but it's not the equals() method you included in your code here, because this one belong to Rectangle3 while the equals() you are testing here should be defined in Box3 and Cube3
NB: Please notice as per assylias's comment, that because b is a boolean there is no need to use if (b == true), just if (b) will be sufficient
It is not the equals function. The line
boolean b = two.equals(four)
Is illegal. It is not within any method and it references variables declared in main()!
I'm in a beginning programming class, and a lot of this had made sense to me up until this point, where we've started working with methods and I'm not entirely sure I understand the "static," "void," and "return" statements.
For this assignment in particular, I thought I had it all figured out, but it says it "can not find symbol histogram" on the line in the main method, although I'm clearly returning it from another method. Anyone able to help me out?
Assignment: "You see that you may need histograms often in writing your programs so you decide for this program to use your program 310a2 Histograms. You may add to this program or use it as a class. You will also write a class (or method) that will generate random number in various ranges. You may want to have the asterisks represent different values (1, 100, or 1000 units). You may also wish to use a character other than the asterisk such as the $ to represent the units of your graph. Run the program sufficient number of times to illustrate the programs various abilities.
Statements Required: output, loop control, decision making, class (optional), methods.
Sample Output:
Sales for October
Day Daily Sales Graph
2 37081 *************************************
3 28355 ****************************
4 39158 ***************************************
5 24904 ************************
6 28879 ****************************
7 13348 *************
"
Here's what I have:
import java.util.Random;
public class prog310t
{
public static int randInt(int randomNum) //determines the random value for the day
{
Random rand = new Random();
randomNum = rand.nextInt((40000 - 1000) + 1) + 10000;
return randomNum;
}
public String histogram (int randomNum) //creates the histogram string
{
String histogram = "";
int roundedRandom = (randomNum/1000);
int ceiling = roundedRandom;
for (int k = 1; k < ceiling; k++)
{
histogram = histogram + "*";
}
return histogram;
}
public void main(String[] Args)
{
System.out.println("Sales for October\n");
System.out.println("Day Daily Sales Graph");
for (int k = 2; k < 31; k++)
{
if (k == 8 || k == 15 || k == 22 || k == 29)
{
k++;
}
System.out.print(k + " ");
int randomNum = 0;
randInt(randomNum);
System.out.print(randomNum + " ");
histogram (randomNum);
System.out.print(histogram + "\n");
}
}
}
Edit: Thanks to you guys, now I've figured out what static means. Now I have a new problem; the program runs, but histogram is returning as empty. Can someone help me understand why? New Code:
import java.util.Random;
public class prog310t
{
public static int randInt(int randomNum) //determines the random value for the day
{
Random rand = new Random();
randomNum = rand.nextInt((40000 - 1000) + 1) + 10000;
return randomNum;
}
public static String histogram (int marketValue) //creates the histogram string
{
String histogram = "";
int roundedRandom = (marketValue/1000);
int ceiling = roundedRandom;
for (int k = 1; k < ceiling; k++)
{
histogram = histogram + "*";
}
return histogram;
}
public static void main(String[] Args)
{
System.out.println("Sales for October\n");
System.out.println("Day Daily Sales Graph");
for (int k = 2; k < 31; k++)
{
if (k == 8 || k == 15 || k == 22 || k == 29)
{
k++;
}
System.out.print(k + " ");
int randomNum = 0;
int marketValue = randInt(randomNum);
System.out.print(marketValue + " ");
String newHistogram = histogram (randomNum);
System.out.print(newHistogram + "\n");
}
}
}
You're correct that your issues are rooted in not understanding static. There are many resources on this, but suffice to say here that something static belongs to a Class whereas something that isn't static belogns to a specific instance. That means that
public class A{
public static int b;
public int x;
public int doStuff(){
return x;
}
public static void main(String[] args){
System.out.println(b); //Valid. Who's b? A (the class we are in)'s b.
System.out.println(x); //Error. Who's x? no instance provided, so we don't know.
doStuff(); //Error. Who are we calling doStuff() on? Which instance?
A a = new A();
System.out.println(a.x); //Valid. Who's x? a (an instance of A)'s x.
}
}
So related to that your method histogram isn't static, so you need an instance to call it. You shouldn't need an instance though; just make the method static:
Change public String histogram(int randomNum) to public static String histogram(int randomNum).
With that done, the line histogram(randomNum); becomes valid. However, you'll still get an error on System.out.print(histogram + "\n");, because histogram as defined here is a function, not a variable. This is related to the return statement. When something says return x (for any value of x), it is saying to terminate the current method call and yield the value x to whoever called the method.
For example, consider the expression 2 + 3. If you were to say int x = 2 + 3, you would expect x to have value 5 afterwards. Now consider a method:
public static int plus(int a, int b){
return a + b;
}
And the statement: int x = plus(2, 3);. Same here, we would expect x to have value 5 afterwards. The computation is done, and whoever is waiting on that value (of type int) receives and uses the value however a single value of that type would be used in place of it. For example:
int x = plus(plus(1,2),plus(3,plus(4,1)); -> x has value 11.
Back to your example: you need to assign a variable to the String value returned from histogram(randomNum);, as such:
Change histogram(randomNum) to String s = histogram(randomNum).
This will make it all compile, but you'll hit one final roadblock: The thing won't run! This is because a runnable main method needs to be static. So change your main method to have the signature:
public static void main(String[] args){...}
Then hit the green button!
For starters your main method should be static:
public static void main(String[] Args)
Instance methods can not be called without an instance of the class they belong to where static methods can be called without an instance. So if you want to call your other methods inside the main method they must also be static unless you create an object of type prog310t then use the object to call the methods example:
public static void main(String[] Args)
{
prog310t test = new prog310t();
test.histogram(1);
}
But in your case you probably want to do:
public static String histogram (int randomNum)
public static void main(String[] Args)
{
histogram(1);
}
Also you are not catching the return of histogram() method in your main method you should do like this:
System.out.print(histogram(randomNum) + "\n");
Or
String test = histogram(randomNum);
System.out.print(test + "\n");
Static methods are part of a class and can be called without an instance but instance methods can only be called from an instance example:
public class Test
{
public static void main(String[] args)
{
getNothingStatic();// this is ok
getNothing(); // THIS IS NOT OK IT WON'T WORK NEEDS AN INSTANCE
Test test = new Test();
test.getNothing(); // this is ok
getString(); // this is ok but you are not capturing the return value
String myString = getString(); // now the return string is stored in myString for later use
}
public void getNothing()
{
}
public static void getNothingStatic()
{
}
public static String getString()
{
return "hello";
}
}
Void means the method is not returning anything it is just doing some processing. You can return primitive or Object types in place of void but in your method you must specify a return if you don't use void.
Before calling histogrom (randomNum) you need to either make histogram static or declare the object that has histogram as a method
e.g
prog310t myClass = new prog310t();
myClass.histogram()
class FDemo {
int x;
FDemo(int i) {
x = i;
}
protected void finalize() {
System.out.println("Finalizing " + x);
}
void generator(int i) {
FDemo o = new FDemo(i);
System.out.println("Creat obj No: " + x); // this line
}
}
class Finalize {
public static void main(String args[]) {
int count;
FDemo ob = new FDemo(0);
for(count=1; count < 100000; count++)
ob.generator(count);
}
}
}
In the line i have commented, the value of x always shows 0(value of x in object ob), why isnt showing the value of object o?? i know if i use o.x i ll be getting the value of x in object o. But still in this code why does it show the value of abject ob rather than object o??
If you want to reference the x in the FDemo you've just created, you should add a getX() function and call that instead of x, like David Wallace said. (I prefer using getters instead of .variable).
Add this to your class:
public int getX(){
return x;
}
And change your problematic line to this:
System.out.println("Creat obj No: " + o.getX());
That should fix it. As a side note, it's considered good practice to explicitly declare whether your variables and methods are private, public or protected.
I have a quick question out of curiosity...if I declare an integer in one method, for example: i = 1, is it possible for me to take that i and use its value in my main class (or another method)? The following code may be helpful in understanding what I'm asking...of course, the code might not be correct depending on what the answer is.
public class main {
public main() {
int n = 1;
System.out.print(n + i);
}
public number(){
i = 1;
}
}
No you cannot! Not unless you make it an instance variable!
Or actually send it to the function as an argument!
First, let's start simple. All methods that are not constructors require a return type. In other words,
public void number(){
i = 1;
}
would be more proper.
Second: the main method traditionally has a signature of public static void main(String[] args).
Now, on to your question at hand. Let's consider a few cases. I will be breaking a few common coding conventions to get my point across.
Case 1
public void number(){
i = 1;
}
As your code stands now, you will have a compile-time error because i is not ever declared. You could solve this by declaring this somewhere in the class. To access this variable, you will need an object of type Main, which would make your class look like this:
public class Main {
int i;
public static void main(String[] args) {
Main myMain = new Main();
myMain.number();
System.out.print(myMain.i);
}
public void number(){
i = 1;
}
}
Case 2
Let's say you don't want to make i a class variable. You just want it to be a value returned by the function. Your code would then look like this:
public class Main {
public static void main(String[] args) {
Main myMain = new Main();
System.out.print(myMain.number());
}
public int number(){ //the int here means we are returning an int
i = 1;
return i;
}
}
Case 3
Both of the previous cases will print out 1 as their output. But let's try something different.
public class Main {
int i = 0;
public static void main(String[] args) {
Main myMain = new Main();
myMain.number();
System.out.print(myMain.i);
}
public void number(){
int i = 1;
}
}
What do you think the output would be in this case? It's not 1! In this case, our output is 0. Why?
The statement int i = 1; in number(), it creates a new variable, also referred to as i, in the scope of number(). As soon as number() finishes, that variable is wiped out. The original i, declared right under public class Main has not changed. Thus, when we print out myMain.i, its value is 0.
Case 4
One more case, just for fun:
public class Main {
int i = 0;
public static void main(String[] args) {
Main myMain = new Main();
System.out.print(myMain.number());
System.out.print(myMain.i);
}
public int number(){
int i = 1;
return i;
}
}
What will the output of this be? It's 10. Why you ask? Because the i returned by number() is the i in the scope of number() and has a value of 1. myMain's i, however, remains unchanged as in Case 3.
You may use a class-scope field to store you variable in a class object or you can return it from one method or pass it as a parameter to the other. Mind that you will need to call your methods in the right order, which is not the best design possible.
public class main {
int n;
int i;
public main() {
n = 1;
System.out.print(n + i);
}
public number(){
i = 1;
}
}
Yes, create a classmember:
public class Main
{
private int i;
public main() {
int n = 1;
System.out.print(n + i);
number();
System.out.print(n + i);
}
public number(){
i = 1;
}
}
void method(){
int i = 0; //has only method scope and cannot be used outside it
}
void method1(){
i = 1; //cannot do this
}
This is because the scope of i is limited to the method it is declared in.
I have been given a piece of code (the class QuestionTwo).
I am asked to state the values of a, b, and c after method mQ2 is invoked on a newly created object of class Q2.
My main.java file
package openuniversity;
public class Main
{
public static void main(String[] args)
{
QuestionTwo qt = new QuestionTwo();
qt.mQ2();
}
}
My QuestionTwo.java class file:
package openuniversity;
public class QuestionTwo
{
int a;
int b = 1;
public void mQ2()
{
{
int c;
int a = 2;
c = a;
}
{
int c;
int a;
c = 3;
a = 4;
}
a++;
}
}
I arrived at:
a: 1
b: 1
c: 3
Note I can also select 'undefined' as an answer?
So would it be 1, 1, undefined as c does not exist outside of the codeblock?
The question:
Study the following code and then select the options from the drop-down lists below that are correct about the values of a, b and c after the method mQ2 is invoked once on a newly created object of class Q2. Note that the answers you choose for a, b and c may or may not be different from each other.
public class Q2
{
int a;
int b = 1;
public void mQ2()
{
{
int c;
int a = 2;
c = a;
}
{
int c;
int a;
c = 3;
a = 4;
System.out.println("c: " + c); //correct place?
}
a++;
}
System.out.println("a: " + a + "b: " + b); // correct place?
}
Since this is homework, I'll restrict my answer to a couple of pointers.
You can verify your solution by printing out the variables after calling mQ2() (hint: you could use System.println() for that).
This is either a trick question or is partially ill-defined (hint: think about which a, b and especially c you're being asked about).
I'd suggest you first print out all the values using System.out.println() after calling mQ2, then step through the code in your mind to try to work out why the values are what they are. Remember that any variable declared is only visible within the scope ({...}s for simplicity), but these variables can have the same name as other variables so they might look like the same even if they're not.
I'd like to particularly point out that c does not exist outside that method.