java/1262: Method with default access can be overridden in another package

[bryce@albatross.co.nz]

>Number:         1262
>Category:       java
>Synopsis:       Method with default access can be overridden in another package
>Confidential:   no
>Severity:       serious
>Priority:       medium
>Responsible:    apbianco
>State:          open
>Class:          sw-bug
>Submitter-Id:   net
>Arrival-Date:   Wed Dec 20 12:18:27 PST 2000
>Closed-Date:    
>Last-Modified:  
>Originator:     Bryce McKinlay
>Release:        unknown-1.0
>Organization:
>Environment:
gcc version 2.96 20000525 (experimental)
>Description:
It is legal in Java for class to declare a method with the
same name as a default (package-private) method in a 
foreign-package superclass.

However, because it is invisible to the sub-class, the
method should NOT override the superclasses method 
declaration.

>How-To-Repeat:
Compile the following code:

// file: pkg1/A.java
package pkg1;
public class A 
{
  void foo () 
  {
    System.out.println ("A.foo()");
  }
  
  public static void main(String[] args) {
    A x = new pkg2.B();
    x.foo();
  }
}

// file: pkg2/B.java
package pkg2;
public class B extends pkg1.A 
{
  protected void foo ()
  {
    System.out.println("B.foo()"); 
  }
}

The correct output (as tested on IBM JDK 1.1.8 and Sun 1.2.2)
should read:

$ java pkg1.A
A.foo()

The gcj-compiled output is:

$ ./pkg
B.foo()
>Fix:

>Release-Note:
>Audit-Trail:

Formerly PR gcj/244

>Unformatted: