C++ syntax err, abstracting static and member function into one templated class
sorry for the title, i don't know what suitable title of this topic. I have this test code http://ideone.com/V8h5K which as compile error, and I don't know what's wrong with this code:
my intention is to create an abstraction so that I can create a container of generic function pointer which don't care if the function pointer is a pointer to static function or a member function.
#include <stdio.h>
#include <vector>
template<typename R=void,
typename A=void,
typename F=R (*)(A)> class Method {
protected: F method;
public : Method(F methodPtr):method(methodPtr){ };
virtual R operator()(A argument) {
开发者_开发知识库 return this->method(argument);
};
typedef F FuncType;
};
template<typename A, typename F> class Method<void,A,F> {
protected: F method;
public : Method(F methodPtr) :method(methodPtr) { };
virtual void operator()(A argument) {
this->method(argument);
};
typedef F FuncType;
};
template<typename R, typename F> class Method<R,void,F> {
protected: F method;
public : Method(F methodPtr) :method(methodPtr) { };
virtual R operator()() {
return this->method();
};
typedef F FuncType;
};
template<typename F> class Method<void,void,F> {
protected: F method;
public : Method(F methodPtr) :method(methodPtr) { };
virtual void operator()() {
this->method();
};
typedef F FuncType;
};
template<typename C=void,
typename R=void,
typename A=void,
typename F=R (C::*)(A)>
class ClassMethod : public Method<R,A,F> {
protected: C& owner;
public : ClassMethod(C& methodOwner,F methodPtr)
:Method<R,A,F>(methodPtr),owner(methodOwner){ };
virtual R operator()(A argument) {
return ((this->owner).*(this->method))(argument);
};
typedef F FuncType;
};
template<typename C, typename A, typename F>
class ClassMethod<C,void,A,F>: public Method<void,A,F> {
protected: C& owner;
public : ClassMethod(C& methodOwner,F methodPtr)
:Method<void,A,F>(methodPtr),owner(methodOwner){ };
virtual void operator()(A argument) {
((this->owner).*(this->method))(argument);
};
typedef F FuncType;
};
template<typename C, typename R, typename F>
class ClassMethod<C,R,void,F>: public Method<R,void,F> {
protected: C& owner;
public : ClassMethod(C& methodOwner,F methodPtr)
:Method<R,void,F>(methodPtr),owner(methodOwner){ };
virtual R operator()() {
return ((this->owner).*(this->method))();
};
typedef F FuncType;
};
template<typename C, typename F>
class ClassMethod<C,void,void,F>: public Method<void,void,F> {
protected: C& owner;
public : ClassMethod(C& methodOwner,F methodPtr)
:Method<void,void,F>(methodPtr),owner(methodOwner){ };
virtual void operator()() {
((this->owner).*(this->method))();
};
typedef F FuncType;
};
// ---- implementation -----
template<typename A> class MethodList {
protected:
std::vector< Method<void,A> > methods;
public:
void add(typename Method<void,A>::FuncType fp) {
this->methods.push_back(Method<void,A>(fp));
}
template<class C> void add(typename C& instance,
typename ClassMethod<C,void,A>::FuncType fp) {
this->methods.push_back(ClassMethod<C,void,A>(instance,fp));
}
void invoke(A argument) {
typename std::vector< Method<void,A> >::iterator it;
for(it=this->methods.begin() ; it!=this->methods.end() ; it++) {
(*it)(argument);
}
}
};
void function1(int arg) {
printf("function1(%d)",arg);
}
class Class1 {
public:
void function1(int arg) {
printf("Class1::function1(%d)",arg);
}
};
int main(int argc,char* argv[] )
{
Class1 inst;
MethodList<int> methodList;
methodList.add(function1);
methodList.add<Class1>(inst,&Class1::function1);
methodList.invoke(123);
return 0;
}
This virtual member function will be instantiated whenever the enclosing class is instantiated on most implementations. You don't have to use it for that instantiation to occur:
virtual R operator()(A argument) {
return this->method(argument);
};
However when F
is a R(C::*)(Args)
, the use of this->method
will be ill-formed. So this derivation is wrong:
// will instantiate Method<R, A, F>
class ClassMethod : public Method<R,A,F>
FWIW you don't need to specialize for R
being void
. You can have a return statement in a void function with an expression that has type void
just fine.
// works fine with R being void
virtual R operator()(A argument) {
return static_cast<R>(this->method(argument));
};
WOW, now I see you are also slicing objects in your MethodList
class template. If you don't know what "slicing" is about, I recommend you to read on basic C++ topics first, before writing this sort of code.
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