Re: Overriding methods with lower permission

From:
Pete Becker <pete@versatilecoding.com>
Newsgroups:
comp.lang.c++
Date:
Mon, 28 Jul 2008 14:42:27 -0400
Message-ID:
<2008072814422775249-pete@versatilecodingcom>
On 2008-07-28 14:35:25 -0400, puzzlecracker <ironsel2000@gmail.com> said:

On Jul 28, 1:53??pm, dani...@mail.ru wrote:

Hello there,
I have the following code:

class Base {
public:
?? virtual void f() {cout << "Base::f()" << endl;}
?? virtual void f(int) {cout << "Base::f(int)" << endl;}

};

class Derived : public Base {
?? void f() {cout << "Derived::f()" << endl;}

};

int main() {
?? Base *ptr = new Derived();
?? ptr->f();
?? ptr->f(1);

?? return 0;

}

As you can see, I overridden Base::f() from public, to private access
permission. Being the Java guy that I am, I expected a compiler error.
But to my surprise, the program ran fine, and called the private
version? Is this a compiler bug (I use g++), or is the function
explicitly converted to public? Thank you.


That's allowed in C++. However, you're not allowed to provide more
accessibility than provided by super class, in other words making it a
private member a public in subclass is prohibited under the current
standard.


No, it's not. I think you're confusing this with the rule about using
directives, which are not allowed to increase access.

--
  Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)

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