How to do "derived" type extensions?
 

Suppose class B is a subtype of class A, i.e.,
class B : public A
{
...
};
Then, in C++, a pointer to type B can be treated as a pointer to type
A. So, I can do the following:
B *b;
A *a = b;
This morphism of types is what makes C++ so powerful.

But now how do I do the following kind of morphism? Suppose I have functions
void foo(const B &);
and
void bar(const A &);
I like to be able to declare a function pointer
void (*fp)(const B&);
and make "polymorphic" assignments like
p = foo; // this is ok in C++.
p = bar; // semantically this make sense, but C++ syntax won't
allow this!

I know the last line of code is not valid C++ because the function
signatures are required to be the same. However, on a semantic level,
I should be allowed to substitute calls to foo with calls to bar. This
is what I mean:
B b;
foo(b); // I can substitute the function foo in this line to
bar(b); // the function bar in this line.
So, at least on a semantic level I should be able to declare a
"polymorphic" function pointer that can points to both foo and bar
types.
void (*fp)(const B&) = bar;
fp(b); // here, b is passed as type A to fp since bar accepts
type A argument.

But I can't, because C++ doesn't work this way.

Now, I know the mantra "When in Rome, do as Romans." So, if this is
not a feature of C++, then I should rework a solution that is in C++.
But I'm just curious if anyone has a "good hack" to simulate this type
of polymorphism in C++?

--

-kira

Re: How to do "derived" type extensions?
 

Kira Yamato wrote:

> Suppose class B is a subtype of class A, i.e.,
> class B : public A
> {
> ...
> };
> Then, in C++, a pointer to type B can be treated as a pointer to type
> A. So, I can do the following:
> B *b;
> A *a = b;
> This morphism of types is what makes C++ so powerful.
>
> But now how do I do the following kind of morphism? Suppose I have
> functions
> void foo(const B &);
> and
> void bar(const A &);
> I like to be able to declare a function pointer
> void (*fp)(const B&);
> and make "polymorphic" assignments like
> p = foo; // this is ok in C++.
> p = bar; // semantically this make sense, but C++ syntax won't
> allow this!

Yes.

>
> I know the last line of code is not valid C++ because the function
> signatures are required to be the same. However, on a semantic level,
> I should be allowed to substitute calls to foo with calls to bar. This
> is what I mean:
> B b;
> foo(b); // I can substitute the function foo in this line to
> bar(b); // the function bar in this line.
> So, at least on a semantic level I should be able to declare a
> "polymorphic" function pointer that can points to both foo and bar
> types.
> void (*fp)(const B&) = bar;
> fp(b); // here, b is passed as type A to fp since bar accepts
> type A argument.
>
> But I can't, because C++ doesn't work this way.

Right. You are using pointer to functions which are low level and have
limited semantics vs what you want. You should instead use something like
boost::function<> where you specify the signature of the function as per
valid call not to match identically with the called function (that is, as
long as the called function returned argument is convertible to your
specified returned type for boost::function and as long as the input
arguments are convertible from your specified ones to the ones the function
actually takes then it compiles).

This is basically some template machinery (you store the original function
pointer as a type independent of the type you are suposed to call but when
calling the proper conversions happen).

> Now, I know the mantra "When in Rome, do as Romans." So, if this is
> not a feature of C++, then I should rework a solution that is in C++.
> But I'm just curious if anyone has a "good hack" to simulate this type
> of polymorphism in C++?

No need for a hack, but a nice C++ compile time typesafe solution, something
like boost::function. Here example of what you want:

#include <iostream>
#include <typeinfo>
#include <boost/function.hpp>

struct A {};
struct B: A {};

bool func(A const& arg)
{
std::cout << "called: " << typeid(arg).name() << std::endl;
}

int main()
{
boost::function<int (B const&)> f = func;
f(B());
}

--
Dizzy

Re: How to do "derived" type extensions?
 

Kira Yamato wrote:
> Suppose class B is a subtype of class A, i.e.,
> class B : public A
> {
> ...
> };

>
> But now how do I do the following kind of morphism? Suppose I have
> functions
> void foo(const B &);
> and
> void bar(const A &);
> I like to be able to declare a function pointer
> void (*fp)(const B&);
> and make "polymorphic" assignments like
> p = foo; // this is ok in C++.
> p = bar; // semantically this make sense, but C++ syntax won't
> allow this!


Would it be acceptable to create a function,
void bar(const B &);
and call bar(const A &) from that function?

LR

Re: How to do "derived" type extensions?
 

On Jan 8, 8:17 am, Kira Yamato <kira...@earthlink.net> wrote:
> Suppose class B is a subtype of class A, i.e.,
> class B : public A
> {
> ...
> };
> Then, in C++, a pointer to type B can be treated as a pointer to type
> A. So, I can do the following:
> B *b;
> A *a = b;
> This morphism of types is what makes C++ so powerful.

What perhaps needs to be noted is that an instance of type B is_an
instance of type A.
Its not correct to say that a pointer to a B 'can be treated' as a
ponter to an A.

Now, ask yourself what makes polymorphism work?

>
> But now how do I do the following kind of morphism? Suppose I have functions
> void foo(const B &);
> and
> void bar(const A &);
> I like to be able to declare a function pointer
> void (*fp)(const B&);
> and make "polymorphic" assignments like
> p = foo; // this is ok in C++.
> p = bar; // semantically this make sense, but C++ syntax won't
> allow this!

Of course it doesn't. If it did we'ld all play russian roulette.

>
> I know the last line of code is not valid C++ because the function
> signatures are required to be the same.
> However, on a semantic level,
> I should be allowed to substitute calls to foo with calls to bar. This
> is what I mean:
> B b;
> foo(b); // I can substitute the function foo in this line to
> bar(b); // the function bar in this line.

Yes, absolutely, its not 'should' - you definitely can and do.

> So, at least on a semantic level I should be able to declare a
> "polymorphic" function pointer that can points to both foo and bar
> types.
> void (*fp)(const B&) = bar;
> fp(b); // here, b is passed as type A to fp since bar accepts
> type A argument.
>
> But I can't, because C++ doesn't work this way.

Thankfully so. You are expecting a pointer to a function with a given
signature to accept a function signature of another type. How do you
expect a call for one signature to induce a call to anything viable?
Are you seriously suggesting that a program should silently construct
an instance of B in order to accept a reference to a base class? Thats
ludicrous and insane.

Basicly, the solution is to declare
void (*fp)(const A&);
the parameter of function at the pointer can be any_kind_of_A.

#include <iostream>
#include <typeinfo>

class A
{
public:
virtual ~A() { } // <---- polymorphism
};

class B : public A { };
class C : public B { };

void bar(const A& a)
{
std::cout << typeid(a).name() << std::endl;
}

void foo(const B& b)
{
std::cout << typeid(b).name() << std::endl;
}

int main()
{
void (*fp)(const A&);

fp = &bar;
A a;
fp(a);
B b;
fp(b); // <- b is passed as a B, see output below
C c;
fp(c);

void (*fp_b)(const B&);
fp_b = &foo;
fp_b(b);
// fp_b(a); // error:
// invalid initialization of reference
// of type 'const B&' from expression
// of type A
}

/*
A
B // <- whats this !
C
B
*/

Read the error again. Do you understand the implications if the
program was allowed to generate an instance of B silently?

>
> Now, I know the mantra "When in Rome, do as Romans." So, if this is
> not a feature of C++, then I should rework a solution that is in C++.
> But I'm just curious if anyone has a "good hack" to simulate this type
> of polymorphism in C++?
>

Sorry, hacking is not programming.

Re: How to do "derived" type extensions?
 

On 2008-01-08 09:37:11 -0500, Salt_Peter <pj_hern@yahoo.com> said:

> On Jan 8, 8:17 am, Kira Yamato <kira...@earthlink.net> wrote:
>> Suppose class B is a subtype of class A, i.e.,
>> class B : public A
>> {
>> ...
>> };
>> Then, in C++, a pointer to type B can be treated as a pointer to type
>> A. So, I can do the following:
>> B *b;
>> A *a = b;
>> This morphism of types is what makes C++ so powerful.
>
> What perhaps needs to be noted is that an instance of type B is_an
> instance of type A.
> Its not correct to say that a pointer to a B 'can be treated' as a
> ponter to an A.

I never even define the term "can be treated". So, how can you say it
is incorrect?

> [...]
>> But now how do I do the following kind of morphism? Suppose I have functions
>> void foo(const B &);
>> and
>> void bar(const A &);
>> I like to be able to declare a function pointer
>> void (*fp)(const B&);
>> and make "polymorphic" assignments like
>> p = foo; // this is ok in C++.
>> p = bar; // semantically this make sense, but C++ syntax won't
>> allow this!
>
> Of course it doesn't. If it did we'ld all play russian roulette.

Huh?

>
>>
>> I know the last line of code is not valid C++ because the function
>> signatures are required to be the same.
>> However, on a semantic level,
>> I should be allowed to substitute calls to foo with calls to bar. This
>> is what I mean:
>> B b;
>> foo(b); // I can substitute the function foo in this line to
>> bar(b); // the function bar in this line.
>
> Yes, absolutely, its not 'should' - you definitely can and do.
>
>> So, at least on a semantic level I should be able to declare a
>> "polymorphic" function pointer that can points to both foo and bar
>> types.
>> void (*fp)(const B&) = bar;
>> fp(b); // here, b is passed as type A to fp since bar accepts
>> type A argument.
>>
>> But I can't, because C++ doesn't work this way.
>
> Thankfully so. You are expecting a pointer to a function with a given
> signature to accept a function signature of another type. How do you
> expect a call for one signature to induce a call to anything viable?

But it is not just any arbitrary type. B is a subtype of A. This is
important. It makes the calls compatible in the following sense (in my
comment after the following paragraph).

> Are you seriously suggesting that a program should silently construct
> an instance of B in order to accept a reference to a base class?

No. There is no need to construct an instance of B. Suppose we forced
void (*fp)(const B &) = (void (*)(const B &)) bar;
where I remind you that
void bar(const A &);
Then the call
fp(B());
is equivalent to
bar(B());
which is perfectly ok because B is a subtype of A.

> Thats
> ludicrous and insane.

Maybe it's just because you didn't understand what I was trying to do.
I'm saying this because your program below is not what I was trying to
do.

>
> Basicly, the solution is to declare
> void (*fp)(const A&);
> the parameter of function at the pointer can be any_kind_of_A.
>
> #include <iostream>
> #include <typeinfo>
>
> class A
> {
> public:
> virtual ~A() { } // <---- polymorphism
> };
>
> class B : public A { };
> class C : public B { };
>
> void bar(const A& a)
> {
> std::cout << typeid(a).name() << std::endl;
> }
>
> void foo(const B& b)
> {
> std::cout << typeid(b).name() << std::endl;
> }
>
> int main()
> {
> void (*fp)(const A&);
>
> fp = &bar;
> A a;
> fp(a);
> B b;
> fp(b); // <- b is passed as a B, see output below
> C c;
> fp(c);
>
> void (*fp_b)(const B&);
> fp_b = &foo;
> fp_b(b);
> // fp_b(a); // error:
> // invalid initialization of reference
> // of type 'const B&' from expression
> // of type A
> }
>
> /*
> A
> B // <- whats this !
> C
> B
> */

Your program is not what I'm trying to accomplish. Your polymorphism
is just on the argument of the function. In fact, why did you even
declare pointers to functions fp and fp_b at all? You could've just
used bar and foo directly.

What I was trying to do is to declare a single pointer to functions
that do not have exact signature but are compatible in the sense I
described above, which I will describe to you again:

If I declare a pointer which points to a function that expects type B,
then it is semantically accepted to use in place of this pointer
another function that actually expects only type A. This is because
when you invoke through this function pointer, the argument of type B
can be passed as an argument of type A. Here we assume B is a subtype
of A.

> [...]
> Sorry, hacking is not programming.

Just curious... What you define hacking as?

--

-kira

Re: How to do "derived" type extensions?
 

On 2008-01-08 08:44:33 -0500, dizzy <dizzy@roedu.net> said:

> Kira Yamato wrote:
>
>> Suppose class B is a subtype of class A, i.e.,
>> class B : public A
>> {
>> ...
>> };
>> Then, in C++, a pointer to type B can be treated as a pointer to type
>> A. So, I can do the following:
>> B *b;
>> A *a = b;
>> This morphism of types is what makes C++ so powerful.
>>
>> But now how do I do the following kind of morphism? Suppose I have
>> functions
>> void foo(const B &);
>> and
>> void bar(const A &);
>> I like to be able to declare a function pointer
>> void (*fp)(const B&);
>> and make "polymorphic" assignments like
>> p = foo; // this is ok in C++.
>> p = bar; // semantically this make sense, but C++ syntax won't
>> allow this!
>
> Yes.
>
>>
>> I know the last line of code is not valid C++ because the function
>> signatures are required to be the same. However, on a semantic level,
>> I should be allowed to substitute calls to foo with calls to bar. This
>> is what I mean:
>> B b;
>> foo(b); // I can substitute the function foo in this line to
>> bar(b); // the function bar in this line.
>> So, at least on a semantic level I should be able to declare a
>> "polymorphic" function pointer that can points to both foo and bar
>> types.
>> void (*fp)(const B&) = bar;
>> fp(b); // here, b is passed as type A to fp since bar accepts
>> type A argument.
>>
>> But I can't, because C++ doesn't work this way.
>
> Right. You are using pointer to functions which are low level and have
> limited semantics vs what you want. You should instead use something like
> boost::function<> where you specify the signature of the function as per
> valid call not to match identically with the called function (that is, as
> long as the called function returned argument is convertible to your
> specified returned type for boost::function and as long as the input
> arguments are convertible from your specified ones to the ones the function
> actually takes then it compiles).
>
> This is basically some template machinery (you store the original function
> pointer as a type independent of the type you are suposed to call but when
> calling the proper conversions happen).
>
>> Now, I know the mantra "When in Rome, do as Romans." So, if this is
>> not a feature of C++, then I should rework a solution that is in C++.
>> But I'm just curious if anyone has a "good hack" to simulate this type
>> of polymorphism in C++?
>
> No need for a hack, but a nice C++ compile time typesafe solution, something
> like boost::function. Here example of what you want:
>
> #include <iostream>
> #include <typeinfo>
> #include <boost/function.hpp>
>
> struct A {};
> struct B: A {};
>
> bool func(A const& arg)
> {
> std::cout << "called: " << typeid(arg).name() << std::endl;
> }
>
> int main()
> {
> boost::function<int (B const&)> f = func;
> f(B());
> }

This is neat! I didn't look deeper into it yet; but assuming this does
indeed do compile-time type-checking, then this is some cool stuff.

Can't wait to dig into boost to see how it's done.

Thanks.

--

-kira

Re: How to do "derived" type extensions?
 

On 2008-01-08 09:18:53 -0500, LR <lruss@superlink.net> said:

> Kira Yamato wrote:
>> Suppose class B is a subtype of class A, i.e.,
>> class B : public A
>> {
>> ...
>> };
>
>>
>> But now how do I do the following kind of morphism? Suppose I have functions
>> void foo(const B &);
>> and
>> void bar(const A &);
>> I like to be able to declare a function pointer
>> void (*fp)(const B&);
>> and make "polymorphic" assignments like
>> p = foo; // this is ok in C++.
>> p = bar; // semantically this make sense, but C++ syntax won't
>> allow this!
>
>
> Would it be acceptable to create a function,
> void bar(const B &);

You mean
void foo(const B &);

> and call bar(const A &) from that function?

Why not?

void foo(const B &b)
{
void bar(const A &);
bar(b);
}

--

-kira

Re: How to do "derived" type extensions?
 

On Jan 8, 10:11 am, Kira Yamato <kira...@earthlink.net> wrote:
> On 2008-01-08 09:18:53 -0500, LR <lr...@superlink.net> said:
>
>
>
> > Kira Yamato wrote:
> >> Suppose class B is a subtype of class A, i.e.,
> >> class B : public A
> >> {
> >> ...
> >> };
>
> >> But now how do I do the following kind of morphism? Suppose I have functions
> >> void foo(const B &);
> >> and
> >> void bar(const A &);
> >> I like to be able to declare a function pointer
> >> void (*fp)(const B&);
> >> and make "polymorphic" assignments like
> >> p = foo; // this is ok in C++.
> >> p = bar; // semantically this make sense, but C++ syntax won't
> >> allow this!
>
> > Would it be acceptable to create a function,
> > void bar(const B &);
>
> You mean
> void foo(const B &);

he's referring to the signature.

>
> > and call bar(const A &) from that function?
>
> Why not?
>
> void foo(const B &b)
> {
> void bar(const A &);
> bar(b);
>
> }
>
> --
>
> -kira

because you'ld call foo(const A& a) instead. or better:

#include <iostream>
#include <typeinfo>

class A { public: virtual ~A() {} };
class B : public A { };

template< typename T >
void func(const T& t )
{
std::cout << typeid(t).name() << std::endl;
}

int main()
{
A a;
func(a);
B b;
func(b);

void (*fp)(const A&);
fp = &func<A>;
fp(a);
fp(b);
}

Re: How to do "derived" type extensions?
 

On Jan 8, 10:07 am, Kira Yamato <kira...@earthlink.net> wrote:
> On 2008-01-08 09:37:11 -0500, Salt_Peter <pj_h...@yahoo.com> said:
>
>
>
> > On Jan 8, 8:17 am, Kira Yamato <kira...@earthlink.net> wrote:
> >> Suppose class B is a subtype of class A, i.e.,
> >> class B : public A
> >> {
> >> ...
> >> };
> >> Then, in C++, a pointer to type B can be treated as a pointer to type
> >> A. So, I can do the following:
> >> B *b;
> >> A *a = b;
> >> This morphism of types is what makes C++ so powerful.
>
> > What perhaps needs to be noted is that an instance of type B is_an
> > instance of type A.
> > Its not correct to say that a pointer to a B 'can be treated' as a
> > ponter to an A.
>
> I never even define the term "can be treated". So, how can you say it
> is incorrect?
>
> > [...]
> >> But now how do I do the following kind of morphism? Suppose I have functions
> >> void foo(const B &);
> >> and
> >> void bar(const A &);
> >> I like to be able to declare a function pointer
> >> void (*fp)(const B&);
> >> and make "polymorphic" assignments like
> >> p = foo; // this is ok in C++.
> >> p = bar; // semantically this make sense, but C++ syntax won't
> >> allow this!
>
> > Of course it doesn't. If it did we'ld all play russian roulette.
>
> Huh?
>
>
>
>
>
> >> I know the last line of code is not valid C++ because the function
> >> signatures are required to be the same.
> >> However, on a semantic level,
> >> I should be allowed to substitute calls to foo with calls to bar. This
> >> is what I mean:
> >> B b;
> >> foo(b); // I can substitute the function foo in this line to
> >> bar(b); // the function bar in this line.
>
> > Yes, absolutely, its not 'should' - you definitely can and do.
>
> >> So, at least on a semantic level I should be able to declare a
> >> "polymorphic" function pointer that can points to both foo and bar
> >> types.
> >> void (*fp)(const B&) = bar;
> >> fp(b); // here, b is passed as type A to fp since bar accepts
> >> type A argument.
>
> >> But I can't, because C++ doesn't work this way.
>
> > Thankfully so. You are expecting a pointer to a function with a given
> > signature to accept a function signature of another type. How do you
> > expect a call for one signature to induce a call to anything viable?
>
> But it is not just any arbitrary type. B is a subtype of A. This is
> important. It makes the calls compatible in the following sense (in my
> comment after the following paragraph).
>
> > Are you seriously suggesting that a program should silently construct
> > an instance of B in order to accept a reference to a base class?
>
> No. There is no need to construct an instance of B. Suppose we forced
> void (*fp)(const B &) = (void (*)(const B &)) bar;
> where I remind you that
> void bar(const A &);
> Then the call
> fp(B());
> is equivalent to
> bar(B());
> which is perfectly ok because B is a subtype of A.
>
> > Thats
> > ludicrous and insane.
>
> Maybe it's just because you didn't understand what I was trying to do.
> I'm saying this because your program below is not what I was trying to
> do.
>
>
>
>
>
> > Basicly, the solution is to declare
> > void (*fp)(const A&);
> > the parameter of function at the pointer can be any_kind_of_A.
>
> > #include <iostream>
> > #include <typeinfo>
>
> > class A
> > {
> > public:
> > virtual ~A() { } // <---- polymorphism
> > };
>
> > class B : public A { };
> > class C : public B { };
>
> > void bar(const A& a)
> > {
> > std::cout << typeid(a).name() << std::endl;
> > }
>
> > void foo(const B& b)
> > {
> > std::cout << typeid(b).name() << std::endl;
> > }
>
> > int main()
> > {
> > void (*fp)(const A&);
>
> > fp = &bar;
> > A a;
> > fp(a);
> > B b;
> > fp(b); // <- b is passed as a B, see output below
> > C c;
> > fp(c);
>
> > void (*fp_b)(const B&);
> > fp_b = &foo;
> > fp_b(b);
> > // fp_b(a); // error:
> > // invalid initialization of reference
> > // of type 'const B&' from expression
> > // of type A
> > }
>
> > /*
> > A
> > B // <- whats this !
> > C
> > B
> > */
>
> Your program is not what I'm trying to accomplish. Your polymorphism
> is just on the argument of the function. In fact, why did you even
> declare pointers to functions fp and fp_b at all? You could've just
> used bar and foo directly.
>
> What I was trying to do is to declare a single pointer to functions
> that do not have exact signature but are compatible in the sense I
> described above, which I will describe to you again:
>
> If I declare a pointer which points to a function that expects type B,
> then it is semantically accepted to use in place of this pointer
> another function that actually expects only type A. This is because
> when you invoke through this function pointer, the argument of type B
> can be passed as an argument of type A. Here we assume B is a subtype
> of A.
>
> > [...]
> > Sorry, hacking is not programming.
>
> Just curious... What you define hacking as?
>

I'm sorry if i upset you, was not my intention. honest.

Re: How to do "derived" type extensions?
 

On Jan 8, 10:07 am, Kira Yamato <kira...@earthlink.net> wrote:
> On 2008-01-08 09:37:11 -0500, Salt_Peter <pj_h...@yahoo.com> said:
>
>
>
> > On Jan 8, 8:17 am, Kira Yamato <kira...@earthlink.net> wrote:
> >> Suppose class B is a subtype of class A, i.e.,
> >> class B : public A
> >> {
> >> ...
> >> };
> >> Then, in C++, a pointer to type B can be treated as a pointer to type
> >> A. So, I can do the following:
> >> B *b;
> >> A *a = b;
> >> This morphism of types is what makes C++ so powerful.
>
> > What perhaps needs to be noted is that an instance of type B is_an
> > instance of type A.
> > Its not correct to say that a pointer to a B 'can be treated' as a
> > ponter to an A.
>
> I never even define the term "can be treated". So, how can you say it
> is incorrect?
>
> > [...]
> >> But now how do I do the following kind of morphism? Suppose I have functions
> >> void foo(const B &);
> >> and
> >> void bar(const A &);
> >> I like to be able to declare a function pointer
> >> void (*fp)(const B&);
> >> and make "polymorphic" assignments like
> >> p = foo; // this is ok in C++.
> >> p = bar; // semantically this make sense, but C++ syntax won't
> >> allow this!
>
> > Of course it doesn't. If it did we'ld all play russian roulette.
>
> Huh?
>
>
>
>
>
> >> I know the last line of code is not valid C++ because the function
> >> signatures are required to be the same.
> >> However, on a semantic level,
> >> I should be allowed to substitute calls to foo with calls to bar. This
> >> is what I mean:
> >> B b;
> >> foo(b); // I can substitute the function foo in this line to
> >> bar(b); // the function bar in this line.
>
> > Yes, absolutely, its not 'should' - you definitely can and do.
>
> >> So, at least on a semantic level I should be able to declare a
> >> "polymorphic" function pointer that can points to both foo and bar
> >> types.
> >> void (*fp)(const B&) = bar;
> >> fp(b); // here, b is passed as type A to fp since bar accepts
> >> type A argument.
>
> >> But I can't, because C++ doesn't work this way.
>
> > Thankfully so. You are expecting a pointer to a function with a given
> > signature to accept a function signature of another type. How do you
> > expect a call for one signature to induce a call to anything viable?
>
> But it is not just any arbitrary type. B is a subtype of A. This is
> important. It makes the calls compatible in the following sense (in my
> comment after the following paragraph).
>
> > Are you seriously suggesting that a program should silently construct
> > an instance of B in order to accept a reference to a base class?
>
> No. There is no need to construct an instance of B. Suppose we forced
> void (*fp)(const B &) = (void (*)(const B &)) bar;
> where I remind you that
> void bar(const A &);
> Then the call
> fp(B());
> is equivalent to
> bar(B());
> which is perfectly ok because B is a subtype of A.
>
> > Thats
> > ludicrous and insane.
>
> Maybe it's just because you didn't understand what I was trying to do.
> I'm saying this because your program below is not what I was trying to
> do.
>
>
>
>
>
> > Basicly, the solution is to declare
> > void (*fp)(const A&);
> > the parameter of function at the pointer can be any_kind_of_A.
>
> > #include <iostream>
> > #include <typeinfo>
>
> > class A
> > {
> > public:
> > virtual ~A() { } // <---- polymorphism
> > };
>
> > class B : public A { };
> > class C : public B { };
>
> > void bar(const A& a)
> > {
> > std::cout << typeid(a).name() << std::endl;
> > }
>
> > void foo(const B& b)
> > {
> > std::cout << typeid(b).name() << std::endl;
> > }
>
> > int main()
> > {
> > void (*fp)(const A&);
>
> > fp = &bar;
> > A a;
> > fp(a);
> > B b;
> > fp(b); // <- b is passed as a B, see output below
> > C c;
> > fp(c);
>
> > void (*fp_b)(const B&);
> > fp_b = &foo;
> > fp_b(b);
> > // fp_b(a); // error:
> > // invalid initialization of reference
> > // of type 'const B&' from expression
> > // of type A
> > }
>
> > /*
> > A
> > B // <- whats this !
> > C
> > B
> > */
>
> Your program is not what I'm trying to accomplish. Your polymorphism
> is just on the argument of the function. In fact, why did you even
> declare pointers to functions fp and fp_b at all? You could've just
> used bar and foo directly.
>
> What I was trying to do is to declare a single pointer to functions
> that do not have exact signature but are compatible in the sense I
> described above, which I will describe to you again:
>
> If I declare a pointer which points to a function that expects type B,
> then it is semantically accepted to use in place of this pointer
> another function that actually expects only type A. This is because
> when you invoke through this function pointer, the argument of type B
> can be passed as an argument of type A. Here we assume B is a subtype
> of A.
>
> > [...]
> > Sorry, hacking is not programming.
>
> Just curious... What you define hacking as?
>
> --
>
> -kira


Not meant to upset you. Honest.
take a look at the following, perhaps you'll get my drift:

#include <iostream>
#include <typeinfo>

class A
{
public:
virtual ~A() {}
virtual void foo() const { std::cout << "A::foo()\n"; }
virtual void bar() const { std::cout << "A::bar()\n"; }
};

class B : public A
{
public:
void foo() const { std::cout << "B::foo()\n"; }
void bar() const { std::cout << "B::bar()\n"; }
};

template< typename T >
void func(const T& t )
{
std::cout << typeid(t).name() << std::endl;
t.foo();
}

int main()
{
A a;
func(a);
B b;
func(b);

void (*fp)(const A&);
fp = &func<A>;
fp(a);
fp(b);
}

/*
A
A::foo()
B
B::foo()
A
A::foo()
B
B::foo()
*/