Re: How to do "derived" type extensions?

From:
Kira Yamato <kirakun@earthlink.net>
Newsgroups:
comp.lang.c++
Date:
Wed, 9 Jan 2008 03:04:52 -0500
Message-ID:
<2008010903045216807-kirakun@earthlinknet>
On 2008-01-08 11:18:23 -0500, Salt_Peter <pj_hern@yahoo.com> said:

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);
}


But the following will not compile:
    fp = &func<B>;

So, if I specialize for func<B>, your code won't work as is.

However, there is probably a solution using templates and casts, as a
few other posters have hinted.

--

-kira

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