CODEWITHSUNDEEP
java
user2287094
user2287094

Reputation: 277

Java: Why are accessible non-static variables from a static methods outside from the class?

In Java when we want to access a non-static variable from a static method, we get a compile error:

"Cannot make a static reference to the non-static field nonStatic"

But I made the other class with a static method, and I could reach that non-static variable. Why is this?

So in class A, the line "return nonStatic" and the line "nonStatic=4" is bad. But in the class App method m, and the class B method m is working.

public class App 
{
    public static void main( String[] args )
    {

        A a = new A();
        m( a  );

        System.out.println(a.nonStatic);

    }

    static void m( A a ) {
        a.nonStatic = 12;  //its good...why?

    }
}

class A{

    int nonStatic = 7; 

    static int getOrSetNonStatic(){
        // return nonStatic;  //error
        // nonStatic = 4;     //error
        return 0;
    }

}

class B {

     static void m( A a ) {
            a.nonStatic = 12;   //its good...why?

        }
}

Upvotes: 1

Views: 135

Answers (2)

ernest_k
ernest_k

Reputation: 45309

What makes the difference is not the fact that nonStatic is accessed from a different class.

Try doing that in A and it will compile:

class A {
    int nonStatic = 7; 
     static void m( A a ) {
            a.nonStatic = 12;
     }
}

So what makes the difference? It's what you read the static variable on:

  • a.nonStatic accesses nonStatic on an instance. Regardless of where this code is, it is OK.
  • Simply typing nonStatic in an instance (non-static) method inside A is similar to a.nonStatic (i.e., this.nonStatic). This is allowed.
  • Simply typing nonStatic in a static method inside A is equivalent to A.nonStatic, which is a problem regardless of where it's written. nonStatic is an instance field, so it cannot be accessed statically (as A.nonStatic anywhere, or as nonStatic in a static method of A)

Upvotes: 2

Nathan Hughes
Nathan Hughes

Reputation: 96385

Because you passed it into the method as an argument.

Maybe it would be clearer if you gave the method parameter a different name. Then you would see that the member field wasn’t visible as a but only under the parameter name.

Upvotes: 2

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