Solved

easy function pointer question

Posted on 2007-03-22
6
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Last Modified: 2010-04-01
Hi,

I have two classes, I want to know how one class can set a pointer to a function to another like:

class A {
    void DoSomething();
    void SetFunctionPointer(?); { m_pFunctionPointer = ?; }
    void* m_pFunctionPointer;
};

class B {
     A* m_p;
     void SayHello() { cout << "Hello" << endl; }
};

void A::DoSomething()
{
    m_pFunctionPointer();
}

B::B()
{
    m_p->SetFunctionPointer(&SayHello());
}

Thanks
0
Comment
Question by:DJ_AM_Juicebox
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6 Comments
 
LVL 86

Accepted Solution

by:
jkr earned 500 total points
ID: 18772052
If you want to call a class method, you need to keep in mind that you can only use it with an instance of a class. E.g.

#include <iostream>
using namespace std;

class B; // fwd. decl.

class A {
public:
    void DoSomething();
    void SetFunctionPointer(void (B::*ptr)()) { m_pFunctionPointer = ptr; }
    void (B::*m_pFunctionPointer)();
};

class B {
public:
     B();
     A* m_p;
     void SayHello() { cout << "Hello" << endl; }
};

void A::DoSomething()
{
    B b;
    (b.*m_pFunctionPointer)();
}

B::B()
{
    m_p->SetFunctionPointer(B::SayHello);
}

void main () {
  A a;
  a.DoSomething();

}

works for that.

0
 
LVL 86

Expert Comment

by:jkr
ID: 18772068
BTW, see also http://www.newty.de/fpt/index.html ("The Function Pointer Tutorials") for more info.
0
 
LVL 39

Expert Comment

by:itsmeandnobodyelse
ID: 18772237
The problem with member function pointers is that you need both the function pointer and an instance of the class the function is a member from. Because of that there are only a few known senseful samples where any one has used member function pointers, You see it in jkr's sample above. He creates a B object before calling the function pointer but the B is any B and has only default initialization. So that B object cannot do any specific things but only that which are valid for all instances of class B. for that purpose you can use static member functions which can be called without needing an instance of B but only with the class prefix:

class B {
public:
     B();
     static  void SayHello() { cout << "In B: Hello" << endl; }
};

You can call it everywhere by simply

         B::SayHello();

And function pointers to static member functions can be treated like function pointers to global functions:

// a typedef makes it simpler to define the type
typedef void (*SayFunc)();

class A
{
      SayFunc  funcPtr;   // we use our typedef here
public:
      A(SayFunc sf) :  funcPtr(sf) { }   // pass it to the constructor        
      void say() {  funcPtr(); }
};

That's all.

Regards, Alex
0
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LVL 53

Expert Comment

by:Infinity08
ID: 18772389
Try this :

#include <iostream>

using namespace std;

class A;

class B {
  public:
    A* m_p;
    B();
    void SayHello() { cout << "Hello" << endl; }
    ~B();
};

class A {
  private:
    void (B::*m_pFunctionPointer)();
  public:
    void DoSomething();
    void SetFunctionPointer(void (B::*fun_ptr)()) { m_pFunctionPointer = fun_ptr; }
};

void A::DoSomething() {
  B b;
  (b.*m_pFunctionPointer)();
}

B::B() {
  m_p = new A();
  m_p->SetFunctionPointer(&B::SayHello);
}

B::~B() {
  delete m_p;
}

int main(void) {
  B b;
  b.m_p->DoSomething();
  return 0;
}
0
 
LVL 53

Expert Comment

by:Infinity08
ID: 18772404
Oops, should have checked before posting ... that's the trouble when you start writing a reply, and get distracted half way through :)
0
 
LVL 11

Expert Comment

by:DeepuAbrahamK
ID: 18777786
Hey DJ,

Try this one: its tidy!
http://www.codeproject.com/cpp/Functor.asp


Best Regards,
DeepuAbrahamK
0

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