CODEWITHSUNDEEP
c++functionpointerscallbackmember
user346443
user346443

Reputation: 4862

C++ call back system. Pointers to member functions

Im trying to create a call back by passing a pointer to member functions but am running into all types of issues.

How can i go about implementing such a thing

template<class T> class A{

   void (T::*callBackFunction)(int);

   public:
       void addCallBack(void (T::*callBackFunction)(int)){
            void (T::*callBackFunction)(int) = &callBackFunction;
       }

       void performCallBack(){ //I want this to activate the specified function in class B
            callBackFunction(3);
       }
};





Class B{

   A<B> a1;
   A<B> a2;

   B(){
       a1.addCallBack(&a1CallBack);
       a2.addCallBack(&a2CallBack);
   }

   void a1CallBack(int i){
       // do stuff
   }

   void a2CallBack(int i){
       // do stuff
   }
};

Upvotes: 0

Views: 2238

Answers (4)

CashCow
CashCow

Reputation: 31445

If what you really want is a callback function that takes int as a parameter and returns void, you can use boost::function.

#include <boost/function.hpp>
#include <boost/bind.hpp>
#include <iostream>

struct A
{
   typedef boost::function1<int, void> func_type;
   func_type func;

   explicit A( func_type func_ ) :
      func( func_ )
   {
   }

   void call( int x )
   {
       func(x);
   }
};

struct B
{
   int myVal;
   explicit B(int val ) : myVal( val ) {}

   void mycall(int x)
    {
       std::cout << (myVal + x) << std::endl;
    }
};

int main()
{
   B b(5);
   A a( boost::bind( &B::mycall, b, _1 ) );
   a.call( 3 );
}

and with any luck it will print 8.

Upvotes: 0

Chris Becke
Chris Becke

Reputation: 36131

You could do something like this. The virtual base class BaseCB allows B to be totally unaware of the type of C but still invoke the callback.

class BaseCB{
public:
  virtual void operator()(int x)=0;
};

template<class ClassT>
class CCallback : public BaseCB {
public:
  typedef void(ClassT::* FuncT)(int);
  FuncT _fn;
  ClassT* _c;
  CCallback(ClassT* c, FuncT fn):_fn(fn),_c(c){}
  void operator()(int x){
    return (_c->*_fn)(x);
  }
};

class B {
public:
  BaseCB* a;
  void invokecb(int n){
    (*a)(n);
  }
};

class C {
  B b;
public:
  C(){
    b.a = new CCallback<C>(this,&C::cb);
    b.invokecb(3);
  };
  void cb(int n)
  {
    cout << n;
  }
};

C c; // create an instance of c and test the cb mechanism

Upvotes: 3

BЈовић
BЈовић

Reputation: 64283

You need the reference (or a pointer) to the object on which you want to call the method.

I would suggest to create a template base class (template parameters are function return value and parameters), and a template derived class (all base template parameters, plus the object type). The derived class holds the reference to the object and the pointer to the member function.

EDIT:

Maybe something like this (not tested) :

template < typename R, typename T1 >
class CbBase
{
  public:
    virtual ~CbBase(){}
    virtual R call( T1 t1 ) = 0;
};

template < typename O, typename R, typename T1 >
class MemFunPtrCb : public CbBase< R, T1 >
{
  public:
    MemFunPtrCb( O &obj_, R (O::*memFcPtr_)( T1 t1 ) ) :
      obj( obj_ ),
      memFcPtr( memFcPtr_ ),
    {
    }

    virtual R call( T1 t1 )
    {
        obj.memFcPtr( t1 );
    }

  private:
    O &obj;
    R (O::*memFcPtr)( T1 t1 );
};

This example is missing :

  • a function to create MemFuncPtrCb object, this function needs to take a reference to the object, and member function, and return shared_ptr to the appropriate base class

  • typedefs

Upvotes: 1

Bj&#246;rn Pollex
Bj&#246;rn Pollex

Reputation: 76876

Pointers to member function require a pointer to an instance as their first parameter. Read this. The general way to do this is to have the callback function take an extra parameter (usually a void *) that can represent extra information or context for the callback handler. This could be the pointer to an object to which the callback would be delegated.

Upvotes: 1

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