Java instance variables

Question

I am learning Java using Codecademy. I am trying to write code that would make an object (in this case a 3 year old dog named Spike) to run on on n legs where n = age. I tried "brute force debugging", but that didn't work. How should I change the line "how to change this" to make the code work?

class Dog {
    public Dog(int dogsAge) {
        int age = dogsAge;
    }

    public void bark() {
        System.out.println("Woof!");
    }

    public void run(int feet) {
        System.out.println("Your dog ran " + feet + " feet!");
    }

    public static void main(String[] args) {
        Dog spike = new Dog(3);
        spike.bark();
        spike.run(this.age); // How to change this?
    }
}

By declaring the variable age in the constructor, it's only visible within the constructor. This is an example of scope. — Drew Kennedy
– Drew Kennedy, Commented Apr 2, 2016 at 19:00

Chunko · Accepted Answer · 2016-04-02 18:13:24Z

3

Your class needs to - add storage for the age variable - set that value in the construcor - provide a way for your main function to access it

class Dog {
 private int age; //storage for the age value within the instance
 public Dog(int dogsAge){
   this.age = dogsAge; // set the value
 }
 // provide a way to access the age
 public int getAge() {
   return this.age;
 }
 public void bark(){
   System.out.println("Woof!");

 }
 public void run(int feet) {
   System.out.println("Your dog ran " + feet + " feet!");
 }


 public static void main(String[] args) {
  Dog spike = new Dog(3);
  spike.bark();
  spike.run(spike.getAge()); // retrieve the age and use it
 }

}

answered Apr 2, 2016 at 18:13

Chunko

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2 Comments

Stepan Over a year ago

What happens to a variable dogsAge if I don't add "public int getAge() { return this.age; }" ? Is it completely unaccessible without dogsAge function?

Chunko Over a year ago

In my code above it would not be inaccessible because the main() function is within the Dog class. However if the main() function were in another class, then it would be inaccessible without getAge() because the 'age' property was declared 'private'. You could make it accessible without the getAge() function, by using 'public int age' instead, however that is generally considered poor practise.

Abdelhak · Accepted Answer · 2016-04-02 18:11:59Z

2

Try to create a variable int age and initialize it in constructor like this:

  class Dog {
  int age;
 public Dog(int dogsAge){
    this.age = dogsAge;
  }

answered Apr 2, 2016 at 18:11

Abdelhak

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Comments

Vikrant Kashyap · Accepted Answer · 2016-04-02 18:57:33Z

2

class Dog {
int age;   //your mistake
 public Dog(int dogsAge){
 this.age = dogsAge;
 }
 public void bark() {
    System.out.println("Woof!");
}

public void run(int feet) {
    System.out.println("Your dog ran " + feet + " feet!");
}

public static void main(String[] args) {
    Dog spike = new Dog(3);
    spike.bark();
    spike.run(spike.age); // How to change this?
}

edited Apr 2, 2016 at 18:57

answered Apr 2, 2016 at 18:11

Vikrant Kashyap

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1 Comment

Stepan Over a year ago

Dog.java:25: error: non-static variable this cannot be referenced from a static context spike.run(this.age); // How to change this? ^ 1 error

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