Re: Call a member function only if it exists
On Jan 1, 12:05 pm, Jens Breitbart <jbreitb...@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;
}
Any suggestion how to solve the problem would be highly appreciated.
Regards,
Jens
To do this in general would require C++ to have introspection. Here is
one way that doesn't require inheritance, but still not very elegant:
template<class T> has_f{
enum{value=false};
};
template<class T,bool>
struct caller_imp{
static void f_(T const&){
::f();
}
};
template<class T,bool>
struct caller_imp<T,true>{
static void f_(T const&arg){
arg.f();
}
};
template<class T>
struct caller<T,true>{
void f(T const&arg){
caller_imp<T,has_f<T>::value>::f(arg);
}
};
then for those classes that implement f()
template<>struct has_f<A>{
enum{value=true};
};
Your framework users would only have to specialize the traits class
has_f.
Lance
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