Re: Call a member function only if it exists
In article
<80d4aa91-f16a-4772-a622-b184fcc63220@v4g2000hsf.googlegroups.com>,
Jens Breitbart <jbreitbart@gmail.com> wrote:
Hi,
I wrote a framework, which uses callback functions. Currently all
classes used with the framework have to implement the callback
function. I would like to define a default behavior for all classes,
that do not implement the function. I am aware that this could easily
be done, by defining a base class and require that all classes used
with the framework inherit for this base class, but I would prefer to
do this without inheritance. The code I have in mind looks similar to
the one below, but I failed to write the caller template.
void f () {
// default behavior
}
struct A {
void f () {
// special behavior for objects of type A
}
};
struct B {
};
int main () {
A a;
B b;
//caller<A>::f(a); // should call a.f()
//caller<B>::f(b); // should call f()
return 0;
}
Something like the following ???
#include <iostream>
// an empty base class
// used if none exists for f_base
struct None{};
// a struct that inherits what the derived class
// inherits from [None can be used if none really
// exists]
template <class Base>
struct f_base:Base
{
void f()
{
std::cout << "Default behavior\n";
}
};
// helper template avoids typing type passed.
template <class T>
void call_f(T &a) { a.f();}
// end of general code.
struct A:f_base<None>
{
};
struct B:f_base<None>
{
void f() {std::cout << "B's behavior\n";}
};
struct C
{
virtual void g()=0;
virtual ~C(){}
};
struct D:f_base<C>
{
void g() {f();}
};
struct E:f_base<C>
{
void f() {std::cout << "E's behavior\n";}
void g() {f();}
};
int main()
{
A a;
B b;
D d;
E e;
C *p = &d;
C *q = &e;
std::cout << "non_virtual calls\n";
call_f(a);
call_f(b);
call_f(d);
call_f(e);
std::cout << "virtual calls\n";
p->g();
q->g();
}
it produces:
non_virtual calls
Default behavior
B's behavior
Default behavior
E's behavior
virtual calls
Default behavior
E's behavior
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