Re: Accessing private member via subclass
Mike Schilling wrote:
public abstract class Super
{
private int i;
void method(Sub s)
{
s.i = 2; // (*)
}
}
public class Sub extends Super
{
}
[...]
However, trying to compile these classes leads to:
Super.java:7: i has private access in Super
s.i = 2;
^
1 error
in both 1.4.2_09 and 1.6.0_06.
Obviously, the error can be removed by changing the line to
((Super)s).i = 2;
And, just as obviously, the error doesn't actually prevent
encapsulation from being broken. For what it's worth, similar (in
fact, almost identical) C# code compiles with no problems.
Any thoughts about this?
Peronally I think the decision to break with a compile-error
is good, because otherwise you migth run into problems if you
decide to declare a member "i" in Sub at a later point of time.
Regards, Lothar
--
Lothar Kimmeringer E-Mail: spamfang@kimmeringer.de
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