class A {
private:
virtual void my_function();
}
class B : public A {
private:
void my_function();
}
Am I allowed to increase the visibility of my_function in classB by
declaring it a public function?
Does the answer depend on how I inherit classA from classB (i.e. public,
protected, private)? 14 1287
s wrote: class A { private: virtual void my_function(); }
class B : public A { private: void my_function(); }
Am I allowed to increase the visibility of my_function in classB by declaring it a public function?
Yes.
Does the answer depend on how I inherit classA from classB (i.e. public, protected, private)?
No.
"s" <youshouldbe@home> wrote in message news:3FFDC248.5070204@home... class A { private: virtual void my_function(); }
class B : public A { private: void my_function(); }
Am I allowed to increase the visibility of my_function in classB by declaring it a public function?
Of course. What do you have in mind that might be stopping you?
Does the answer depend on how I inherit classA from classB (i.e. public, protected, private)?
Yes - you won't be able to access it externally if B uses private
inheritance.
On Thu, 08 Jan 2004 16:08:59 -0500, jeffc wrote: Does the answer depend on how I inherit classA from classB (i.e. public, protected, private)?
Yes - you won't be able to access it externally if B uses private inheritance.
No, classA will not be accessable to users of classB, but that is
irrelevant for the question.
HTH,
M4
"s" <youshouldbe@home> skrev i en meddelelse news:3FFDC248.5070204@home... class A { private: virtual void my_function(); }
class B : public A { private: void my_function(); }
Am I allowed to increase the visibility of my_function in classB by declaring it a public function?
Yes - but I do not see any practical reason to do so. Does the answer depend on how I inherit classA from classB (i.e. public, protected, private)?
No.
Remember to have a virtual destructor in A.
Kind regards
Peter
Rolf Magnus wrote: s wrote:
class A { private: virtual void my_function(); }
class B : public A { private: void my_function(); }
Am I allowed to increase the visibility of my_function in classB by declaring it a public function?
Yes.
Right-o! Does the answer depend on how I inherit classA from classB (i.e. public, protected, private)?
No.
I believe it does. Try this in the standard-compliant (ha) compiler of
your choice:
class A {
virtual void my_function() { }
}
class B: A {
void my_function() { A::myfunction( ); }
}
int main( )
{
A a;
}
"Martijn Lievaart" <m@remove.this.part.rtij.nl> wrote in message
news:pa****************************@remove.this.pa rt.rtij.nl... On Thu, 08 Jan 2004 16:08:59 -0500, jeffc wrote:
Does the answer depend on how I inherit classA from classB (i.e.
public, protected, private)?
Yes - you won't be able to access it externally if B uses private inheritance.
No, classA will not be accessable to users of classB, but that is irrelevant for the question.
He asked if the answer depends on how he inherits classA from classB. The
answer is yes. classA will not be accessible to users of classB.
"jeffc" <no****@nowhere.com> wrote in message
news:3f********@news1.prserv.net... "Martijn Lievaart" <m@remove.this.part.rtij.nl> wrote in message news:pa****************************@remove.this.pa rt.rtij.nl... On Thu, 08 Jan 2004 16:08:59 -0500, jeffc wrote:
> Does the answer depend on how I inherit classA from classB (i.e. public,> protected, private)?
Yes - you won't be able to access it externally if B uses private inheritance.
No, classA will not be accessable to users of classB, but that is irrelevant for the question.
He asked if the answer depends on how he inherits classA from classB. The answer is yes. classA will not be accessible to users of classB.
Correction - if the function remains private in classA, then it doesn't
matter.
On Fri, 09 Jan 2004 07:42:31 -0500, Jeff Schwab wrote: Does the answer depend on how I inherit classA from classB (i.e. public, protected, private)?
No.
I believe it does. Try this in the standard-compliant (ha) compiler of your choice:
class A { virtual void my_function() { } }
class B: A { void my_function() { A::myfunction( ); } }
int main( ) { A a; }
Your point? You cannot call a member of a private base class, but how does
that relate to the issue at hand?
M4
Jeff Schwab wrote: Rolf Magnus wrote: s wrote:
class A { private: virtual void my_function(); }
class B : public A { private: void my_function(); }
Am I allowed to increase the visibility of my_function in classB by declaring it a public function?
Yes.
Right-o!
Does the answer depend on how I inherit classA from classB (i.e. public, protected, private)?
No.
I believe it does. Try this in the standard-compliant (ha) compiler of your choice:
class A { virtual void my_function() { } }
class B: A { void my_function() { A::myfunction( ); }
You call A::my_function (I assume you have a typo there), which is
private in A from a member function of B, which is of course not
possible, but that doesn't have anything to do with overriding the
virtual function. B can't call A::my_function, but A or a friend of A
could, and the polymorphism would still work.
}
int main( ) { A a; }
jeffc wrote: "Martijn Lievaart" <m@remove.this.part.rtij.nl> wrote in message news:pa****************************@remove.this.pa rt.rtij.nl... On Thu, 08 Jan 2004 16:08:59 -0500, jeffc wrote:
>> Does the answer depend on how I inherit classA from classB (i.e. >> public, protected, private)? > > Yes - you won't be able to access it externally if B uses private > inheritance. No, classA will not be accessable to users of classB, but that is irrelevant for the question.
He asked if the answer depends on how he inherits classA from classB.
Right.
The answer is yes.
No, it isn't.
classA will not be accessible to users of classB.
Right, it won't. But that wasn't the question. classB does have a member
function my_function, and that function has lower access restrictions
than the one in A, and it does override A::my_function.
On Fri, 09 Jan 2004 10:36:55 -0500, jeffc wrote: "jeffc" <no****@nowhere.com> wrote in message news:3f********@news1.prserv.net... "Martijn Lievaart" <m@remove.this.part.rtij.nl> wrote in message news:pa****************************@remove.this.pa rt.rtij.nl... > On Thu, 08 Jan 2004 16:08:59 -0500, jeffc wrote: > > >> Does the answer depend on how I inherit classA from classB (i.e. public, > >> protected, private)? > > > > Yes - you won't be able to access it externally if B uses private > > inheritance. > > No, classA will not be accessable to users of classB, but that is > irrelevant for the question.
He asked if the answer depends on how he inherits classA from classB. The answer is yes. classA will not be accessible to users of classB.
Correction - if the function remains private in classA, then it doesn't matter.
It still doesn't matter anyhow, the writer of classB is allowed to
give the virtual any access he wants, it does not depend on the type of
inheritance.
M4
"Martijn Lievaart" <m@remove.this.part.rtij.nl> wrote in message
news:pa****************************@remove.this.pa rt.rtij.nl... Correction - if the function remains private in classA, then it doesn't matter.
It still doesn't matter anyhow, the writer of classB is allowed to give the virtual any access he wants, it does not depend on the type of inheritance.
What I was getting at (and I know it's irrelevant now) is the following:
class A
{
public:
void f();
};
class B : public A
{
};
B().f(); // this works
class B : private A
{
};
B().f(); // this does NOT work, so it matters if B publicly or privately
inherits from A
Martijn Lievaart wrote: On Fri, 09 Jan 2004 07:42:31 -0500, Jeff Schwab wrote:
Does the answer depend on how I inherit classA from classB (i.e. public, protected, private)?
No.
I believe it does. Try this in the standard-compliant (ha) compiler of your choice:
class A { virtual void my_function() { } }
class B: A { void my_function() { A::myfunction( ); } }
int main( ) { A a; }
Your point? You cannot call a member of a private base class, but how does that relate to the issue at hand?
It doesn't. I saw "visibility" and thought "access."
Rolf Magnus wrote: Jeff Schwab wrote:
Rolf Magnus wrote:
s wrote: class A { private: virtual void my_function(); }
class B : public A { private: void my_function(); }
Am I allowed to increase the visibility of my_function in classB by declaring it a public function?
Yes.
Right-o!
Does the answer depend on how I inherit classA from classB (i.e. public, protected, private)?
No.
I believe it does. Try this in the standard-compliant (ha) compiler of your choice:
class A { virtual void my_function() { } }
class B: A { void my_function() { A::myfunction( ); }
You call A::my_function (I assume you have a typo there), which is private in A from a member function of B, which is of course not possible, but that doesn't have anything to do with overriding the virtual function. B can't call A::my_function, but A or a friend of A could, and the polymorphism would still work.
You are correct. As apology for the typo's, here's a version that
proves your point:
#include <iostream>
class A
{
virtual void my_function( ) { std::cout << "hello\n"; }
public:
void call_my_function( ) { my_function( ); }
};
class B: A
{
public:
virtual void my_function( ) { std::cout << "world\n"; }
};
int main( )
{
B b;
b.my_function( );
} This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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