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Level 12

07.11.2020
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11comments

Hi Coders !

Question about the task

Resolved

1. Why we should use empty class constructor ? 2. I can't understand connection between method and main method ! 3. I'm completely confused. * I know that I ask question a lot but I'm just student.

Implement the boolean fight(Cat anotherCat) method:
implement a mechanism for cats to fight based on their weight, age and strength.
You can decide for yourself how a cat's characteristics factor into combat.
The method should determine whether we (this) won the fight, i.e. return true if we won and false if not.

The following condition must be met:
if cat1.fight(cat2) returns true,
then cat2.fight(cat1) must return false

Requirements:

The Cat class must have a constructor with no parameters.
The Cat class must contain public fields age, weight, and strength.
The Cat class must have a fight method.
In the fight method, implement a mechanism for cats to fight based on their weight, age and strength.
The method should return the same value every time we fight the same cat.
If cat1 beats cat2, then cat2 must lose to cat1.

package com.codegym.task.task05.task0502; /* Implement the fight method */ public class Cat { public int age; public int weight; public int strength; public Cat() { } public boolean fight(Cat anotherCat) { //write your code here if ((cat1.strength >= cat2.strength) && (cat1.weight >= cat2.weight) && (cat1.age >= cat2.age) ) { return true ; } else { return false ; } } public static void main(String[] args) { Cat cat1 = new Cat() ; cat1.age = 5 ; cat1.strength = 9 ; cat1.weight = 5 ; Cat cat2 = new Cat() ; cat2.age = 6 ; cat2.strength = 10 ; cat2.weigth = 6 ; System.out.println(cat1.fight(cat2)); } }

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Ryan White

Level 8 , New Plymouth, New Zealand

29 November 2020, 18:24

I actually found a slightly different way to pass this, which is still correct. I created seperate win conditions as boolean, then used the if statement to check if all conditions were true: public boolean fight(Cat anotherCat) { boolean win = this.age > anotherCat.age; boolean win2 = this.weight > anotherCat.weight; boolean win3 = this.strength > anotherCat.strength; if (win && win2 && win3) return true; else return false;

lavdon

Level 5 , Charleville-Mezieres, France

25 November 2020, 13:51

all you have to do is create the logic in comparing two cats, no need to create cat objects. if((this.age > anotherCat.age) && (this.weight > anotherCat.weight) && (this.strength > anotherCat.strength){ return true; } else return false; }

Switch/Cypher

Level 25 , Bexleyheath, United Kingdom

7 November 2020, 15:51

Main method is the bit of the program that 'runs' on the opening of the program. Other methods are supplemental formulas that do things within the program.

Class Someclass{
public void someMethod(){}
public void someOtherMethod(){}

public static void main{
someMethod;
someOtherMethod;
}
}

The main contains instructions to run the other methods upon the 'running' of the program. You don't have to use an 'empty' class constructor. There is a difference between 'empty' and 'no parameters':

Class SomeClass{
int someInt;
String someString;

SomeClass(int someInt, String someString){
this.someInt = someInt;
this. someString = someString;
}

Here is a constructor with parameters (things that are given to a constructor to do stuff with). What you have in your code is called a 'default constructor' - it just makes an instance of the class (an object of the class/ instantiates the class) and doesn't DO anything other than that.

Switch/Cypher

Level 25 , Bexleyheath, United Kingdom

7 November 2020, 15:51

If you got more questions then post them here and I'll do my best to make things clearer for you.

Шарох Team Lead

7 November 2020, 16:15

Thank you ! So what should i do with my code? Why we should use that cat constructor ? My code doesnt work :(

Switch/Cypher

Level 25 , Bexleyheath, United Kingdom

7 November 2020, 17:56

Oh right. Your code doesn't work because your "fight" method doesn't know what "cat1" and "cat2" are. It can't see them. It can only "see" what is passed to it and what it's called on. That's why you need "this" which is the cat that you are calling the method on. And "anotherCat" which is what the parameter variable is called. When you give the method "cat2" as that variable then "anotherCat" refers to the item you passed within the variable. Here is a version of your code that compiles - I can't promise that it passes the task, you will need to fiddle with it - but this code "works".

public class Cat {
    public int age;
    public int weight;
    public int strength;

    public Cat() { }

    public boolean fight(Cat anotherCat) {

        return (this.strength >= anotherCat.strength) && (this.weight >= anotherCat.weight) && (this.age >= anotherCat.age);
    }

    public static void main(String[] args) {
        Cat cat1 = new Cat();
        cat1.age = 5;
        cat1.strength = 9;
        cat1.weight = 5;
        Cat cat2 = new Cat();
        cat2.age = 6;
        cat2.strength = 10;
        cat2.weight = 6;

        System.out.println(cat1.fight(cat2));
    }
}

Шарох Team Lead

7 November 2020, 19:32

Wow ! I tought using "this" in method is have no meaning . Thank you for your help :) But i still dont understand why method cant see our object?

Switch/Cypher

Level 25 , Bexleyheath, United Kingdom

7 November 2020, 20:15

Because a reference to it is only made in the main method. I can see how at this stage it's confusing and that's because in a 'real' piece of code these two things wouldn't normally be in the same place. The Cat class would be in a separate Cat class file and the main method in a Main class. It's confusing because this particular 'main' is in the same Class. Imagine if I were referring to some totally other class in a different file. Once you progress in the course and start to understand properly what 'static' and 'non-static' mean and how encapsulation can 'hide' certain variables this will all make much more sense. 'Scope' is the magic word here. The reference to cat1 and cat2 isn't within the 'scope' of the fight method and they need to be 'passed' to that method in order to be within the 'scope'. It's a big topic that you will spend a lot more time on, don't worry.

Switch/Cypher

Level 25 , Bexleyheath, United Kingdom

7 November 2020, 20:21

For your interest here is some code that does allow those variables to be 'seen' by the method:

public static class MyCat{
        private int age;
        private int weight;
        private String name;
        private static MyCat cat1;
        private static MyCat cat2;

        public MyCat(int age, int weight, String name) {
            this.age = age;
            this.weight = weight;
            this.name = name;
        }

        public static void fight(){
            if (cat1.age > cat2.age){
                System.out.println("Cat One Wins!");
            }

        }


        public static void main(String[] args) {
            cat1 = new MyCat(23, 54,"Billy");
            cat2 = new MyCat(54,23,"bobby");
            fight();

        }
    }

Don't worry if you don't get it yet, but I've put two static variables in the class, then I just initialise them in the main method. Then the "fight" method can see them.

Шарох Team Lead

9 November 2020, 06:00

Wow ! Beautiful ! Thank you! I gussed that one thing is need in previous code . You are clearly helped me. :)

Volodymyr

Level 9 , L'viv, Ukraine

3 February 2021, 18:19

Hi, Switch/Cyfer. Could you please, explain to me, why "this.strength" belongs to cat1 and "anotherCat.strength" - to cat2. I dont understand "this." points on class fields and "fight" method accepts just "anotherCat" but nwo cats are comparing... Help, I`m stuck.