6.6.5

Example: Default Access Fields, Methods, and Constructors

NAMES

package pointsUser;

class Test {

public static void main(String[] args) {

points.Point p = new points.Point();

System.out.println(p.x + " " + p.y);

}

}

or by using a single type import declaration ( 7.5.1) that mentions the fully

qualfied name, so that the simple name may be used thereafter:

package pointsUser;

import points.Point;

class Test {

public static void main(String[] args) {

Point p = new Point();

System.out.println(p.x + " " + p.y);

}

}

However, this compilation unit cannot use or import

points.PointList

, which

is not declared

public

 and is therefore inaccessible outside package

points

.

6.6.5   Example: Default Access Fields, Methods, and Constructors

If none of the access modifiers

public

,

protected

, or

private

 are specified, a

class member or constructor is accessible throughout the package that contains the

declaration of the class in which the class member is declared, but the class mem 

ber or constructor is not accessible in any other package. If a

public

 class has a

method or constructor with default access, then this method or constructor is not

accessible to or inherited by a subclass declared outside this package.

For example, if we have:

package points;

public class Point {

public int x, y;

void move(int dx, int dy) { x += dx; y += dy; }

public void moveAlso(int dx, int dy) { move(dx, dy); }

}

then a subclass in another package may declare an unrelated

move

 method, with

the same signature ( 8.4.2) and return type. Because the original

move

 method is

not accessible from package

morepoints

,

super

 may not be used:

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