[FIXED] Struggling to understand the effects of access modifiers and the keyword "static"

Issue

I'm pretty new to Java. I've just written a small example to observe the effects of access modifiers. Maybe the code does not make much sense since I'm just testing things out.

The first code snippet looks like this:

class Tet {
    static double a;
    double b;
    public Tet(double a, double b){
        this.a=a;
        this.b=b;
    }
    
    public void get(){
        a=5;
        
    }
    
    public static void main(String[] args){
        Tet tet1 = new Tet(2, 5);
        Tet tet2 = new Tet(4, 5);
        System.out.println(tet1.a);
        a=4;
        System.out.println(tet1.a);
        tet1.get();
        System.out.println(tet1.a);
        System.out.println(a);
        
    }
}

After this, I made some changes to the code and compiled it again:

class Tet {
    static double a;
    double b;
    public Tet(double a, double b){
        this.a=a;
        this.b=b;
    }
    
    public static void get(){
        a=5;
        
    }
    
    public static void main(String[] args){
        Tet tet1 = new Tet(2, 5);
        Tet tet2 = new Tet(4, 5);
        System.out.println(a);
        get();
        System.out.println(tet1.a);
        System.out.println(a);
        System.out.println(tet2.a);
        
    }
}

After it runs, the cmd looks like this: enter image description here

I really don't know where is number 4 coming from.

Solution

1. All Java's variables must be known at compile time unlike Python, in either way:
   1. declare local variable (in this case, declare int a in your main)
   2. qualify variable (in this case, change to Tet.a)
   3. (import variable, but it's a bit longer and redundant this case)
2. You're touching static area - an special storage. static variables are shared across all class instances. So:

public static void main(String[] args){
    Tet tet1 = new Tet(2, 5); // set Tet's `a` to 2
    // update `a` to 4.
    // `a` is shared across `tet1` and `tet2`!
    Tet tet2 = new Tet(4, 5);
    // Watch out, `Tet`'s `a` is shared! It'll output 4
    System.out.println(tet1.a); 
    System.out.println(tet1.a);
    tet1.get(); // `get`, but updates `a` to 5
    System.out.println(tet1.a); // This will be 5 as well
}

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If this answer resolves your question, consider mark this as "resolved" :)

Answered By - user-id-14900042
Answer Checked By - Terry (JavaFixing Volunteer)