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);
}
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