CodeGym/Help with Java Tasks/How to create a copy of another rectangle passed in as an...

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25.04.2019
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How to create a copy of another rectangle passed in as an argument?

Question about the task

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hey this code doesn't work i know it something to do with th way i create a copy of another rectangle passed in as an argument ON LINE 36~.. what am i missing? thanks!

package com.codegym.task.task05.task0520;

/*
Create a Rectangle class

*/


public class Rectangle {
    //write your code here
    int top;
    int left;
    int width;
    int height;

    public Rectangle(int top){
        this.top = top;
        this.left = 01;
    }

    public Rectangle(int top, int left){
        this.top = top;
        this.left = left;
        this.width = 0;
    }


    public Rectangle(int top, int left, int width){
        this.top = top;
        this.left = left;
        this.width = width;
        this.height = width;
    }


    public Rectangle(anotherObj top, anotherObj left){
        this.top = top;
        this.left = left;
    }

    public static void main(String[] args) {
        Rectangle num1 = new Rectangle(2);
        Rectangle num2 = new Rectangle(3,5);
        Rectangle num3 = new Rectangle(num2.top,num2.left);
        System.out.println(num3.top, num3.left);

    }
}

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Guadalupe Gagnon

Level 37 , Tampa, United States

25 April 2019, 15:17

The requirements say to "create a copy of another rectangle passed in as an argument". With this requirement, the code at line 36 should be:

public Rectangle(Rectangle toCopy){
   // code that set the top, left, width, and height
   // to that of the 'toCopy' rectangle
}

Line 44 should use the constructor above.

dean

Level 8 , Deutschland, Germany

25 April 2019, 16:00

so if i got it right, it should be:

public Rectanlge(otherObj top, otherObj left){
this.top = top;
this.left = left;
}

Rectangle test1 = new Rectangle(3,5); // uses a different constructor not shown in this code
Rectangle test2 = new Rectangle(test1.top, test1.left);

is it correct? :))

Guadalupe Gagnon

Level 37 , Tampa, United States

25 April 2019, 16:13

no. The code that I shared is correct. You need to get rid of that constructor and use the one I shared. Also you should have 4 constructors:

public Rectangle(int left, int top, int width, int height){
   // 4 parameters are specified: left, top, width, height
}

public Rectangle(int left, int top){
   // width/height is not specified (both are 0)
}

public Rectangle(int left, int top, int width){
   // height is not specified (it is equal to the width), we'll create a square
}

public Rectangle(Rectangle toCopy){
   //create a copy of another rectangle passed in as an argument
}

Also, you do not need to use the 'this' keyword when there are no naming conflicts. For example:

public Rectangle(int top, int left){
   this.top = top;
   // ↑ needed here because class variable and local variable are both 'top'
   this.left = left;
   // ↑ needed here because class variable and local variable are both 'left'
   width = 0;
   // not needed here because there is not local variable named 'width', it
   // will use the class width without you having to specify
}

dean

Level 8 , Deutschland, Germany

25 April 2019, 16:40

Ok thanks but wait what should l put on line 14? im confused a bit with a reference to the other object.. thanks also regarding the name conflict, so basically i can write the code below? i thought it is mandatory to pass the instance reference to those variables attributes.. also i call it from the main method which is static is it possible i think i need to call it from outside the main method or?

public class Bla{
    String str;
     int num;

      public static void main(String[] args){
              Bla bs1 = new Bla("dsa", 43);
      }

     public Bla(String test, int t1){
         str = test;
         num = t1;
    }
}

Guadalupe Gagnon

Level 37 , Tampa, United States

25 April 2019, 17:18

The code I shared above in my last code is supposed to replace lines 16-39 in your code. You can keep lines 10-14 the same. This task does not test main, so you don't need anything in there other than for testing purposes. The reference to the other object, you must mean the constructor that takes another rectangle and copies it? So the code should be this:

public Rectangle(Rectangle toCopy){
   top = toCopy.top;
   left = toCopy.left;
   width = toCopy.width;
   length = toCopy.length;
}

and the code that would call this constructor (in main):

Rectangle first = new Rectangle(1,2,3,4);
Rectangle copy = new Rectangle(first); //passes first rectangle in

So the first line creates a rectangle with the values 1, 2 ,3 ,4 set for top, left, width, height. The second line of code will call the constructor that takes a Rectangle object as the argument and sets the values of top, left, width, and height to that of the Rectangle passed in.

Guadalupe Gagnon

Level 37 , Tampa, United States

25 April 2019, 17:29

As regards to the 'this' keyword: In this case it is only needed when variable names collide. The code that you shared is perfectly valid. You can name variables anything you want, from easy to understand to obscure gibberish. The computer doesn't care either way as long as your code is valid. I would recommend using names that are descriptive because it makes it 10 times harder to debug your code if you are constantly having to figure out what the variables are used for. They actually have contest to create as unreadable code as possible doing simple thing. For example this is Java code that output the value of PI (i think):

Guadalupe Gagnon

Level 37 , Tampa, United States

25 April 2019, 17:29

public class Solution
{
    public static void main(String[] args)
    {long ___;long _____[]=new long[0x10];
        {_____[0]=System.nanoTime();
            ___=System.nanoTime()%9223372036854773232L;
            for(int _______=0;_______<0x10;++_______)
            {_____[_______]=(9223372036854773232L*_____[_______]+___&Long.MAX_VALUE)%9223372036854775807L;
                ___=9223372036854773232L*_____[_______]+___;}}
        int ______=0;long _______=0;long ________=0;
        for(;________<0xFFFFFFL;++________)
        {long _________=(9223372036854773232L*_____[______]+___&Long.MAX_VALUE)%9223372036854775807L;
            ___=9223372036854773232L*_____[______]+___;
            _____[______]=_________;
            double __________=(double)_____[______]/Long.MAX_VALUE;
            ++______;
            if(______>=0x10)
            {______=0;}
            _________=(9223372036854773232L*_____[______]+___&Long.MAX_VALUE)%9223372036854775807L;
            ___=9223372036854773232L*_____[______]+___;
            _____[______]=_________;
            double ___________=(double)_____[______]/Long.MAX_VALUE;
            ++______;
            if(______>=0x10)
            {______=0;}
            if(Math.sqrt(__________*__________+___________*___________)<=1)
            {++_______;}}
        System.out.println(4.*_______/________);}}

dean

Level 8 , Deutschland, Germany

18 May 2019, 04:17

loll crazy obscured code ^^ in this example you've wrote: Rectangle first = new Rectangle(1,2,3,4); Rectangle copy = new Rectangle(first); //passes first rectangle in Does 'copy' and 'first' use the same object reference?

Guadalupe Gagnon

Level 37 , Tampa, United States

18 May 2019, 05:11

No. They will store references to different objects (hence the 'new' keyword). This is called shallow copying, where the data inside each object is the same, but the objects themselves are separate. There is a way to deep copy objects, but that is in much later lessons. So if you take this bit of code:

public static void main(String[] args) {
        Rectangle num1 = new Rectangle(2,5,8);
        Rectangle num2 = num1;
        Rectangle num3 = new Rectangle(num1);

        System.out.println(num1 == num2);
        System.out.println(num1 == num3);
        System.out.println(num2 == num3);
}

lines 2-4 create 3 rectangles. the first one is brand new, the second one is a reference to the first, and the third one is a shallow copy of the first. Then line 6-8 compare each triangle for equality. The first one, comparing 1 and 2, returns true. The next 2, which both compare 1 to three (remember that 2 is just a reference to 1), and return false.

dean

Level 8 , Deutschland, Germany

20 May 2019, 09:49

thanks that really makes sense, just few more questions if you don't mind please :) 0. when it is useful to create objects from another object (aka shallow/deep copies)? 1. What is the code to create multiple objects with the same basic info in a loop? i just dont know how to write the syntax for it.. 2, in the code below, when the code goes to the next string? i mean when the program knows to move to the next var is it one digit/letter per string variable or?

public class MainClass
{
  public static void main(String[] args) throws IOException
  {
    BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    String[] list = new String[10];

    for (int i = 0; i < list.length; i++)
    {
      list[i] = reader.readLine();
     }
  }

thanks again! :)

Guadalupe Gagnon

Level 37 , Tampa, United States

20 May 2019, 13:44

useful

0) This is an advanced topic and the reasons for doing one or the other is going to be on a case by case basis. I don't even know why you would use one versus the other, though some of the later tasks before level 20 have you using deep/shallow copying. Here is an article that explains it a little: https://www.geeksforgeeks.org/deep-shallow-lazy-copy-java-examples/ 1) You can use an array of whatever objects you want to create. String[], int[], Dog[], Human[] 2) Im not sure i understand this question, but I will try to answer: in that code the Array list holds 10 Strings that are accessed by their index number, starting at 0 and going through 9. The loop accesses each index individually and sets it to the value returned by the readline() method, and the readline method will return whatever the uses types until enter is hit. Try this code out, pay attention to the output and see if you can use that to figure out what the code does:

public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        String[] list = new String[4];

        for (int i = 0; i < list.length; i++) {
            System.out.println("Loop i = " + i + ": Calling readLine()");
            System.out.print("\t");
            list[i] = reader.readLine();
            System.out.println("\tReadline returned succesfully, list[" + i + "] set to:");
            System.out.println("\t" + list[i]);
            System.out.println("\tList values: " + Arrays.toString(list));
        }
}

dean

Level 8 , Deutschland, Germany

21 May 2019, 04:06

thanks a lot just to clarify for myself few things: 0. i'm trying to understand the basic concept of using an object in another object, when/what for it is used in real life programs for? 2. so if i got it right by default the 'enter' key indicates that the string/int/object input is completed in the readLine() method? that what i was curious about.. thanks a ton again!

Guadalupe Gagnon

Level 37 , Tampa, United States

21 May 2019, 13:35

0) There are many things that using an object inside another object is good for. The most common is comparisons. Take for example a playing card: It has a rank (2,3,4,5,6,7,8,9,10,J,Q,K,A) and a suit (Club, Spade, Diamond, Heart). You could write a method that takes a card object and compares the rank and returns true/false depending on whether the original card is ranked higher, or the other card is ranked higher. Here is just example code that takes a card and call the fictional compare method on another card. If it returns true then the first card 'wins', else the second one does:

if (card1.compare(card2)) output "Player 1 wins!!"
else output "Player 2 wins"

dean

Level 8 , Deutschland, Germany

21 May 2019, 14:44

thanks thats helpful and make sense also what about my second question above if u know maybe :p thanks!

Guadalupe Gagnon

Level 37 , Tampa, United States

21 May 2019, 15:12

solution

Yes. The readLine() method will allow you to type anything and only returns when you hit enter.