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Member Function Pointers

Started by doctorsixstring Sep 30, 2002 at 10:40 AM 2 replies 200+ views

Original Post

Sep 30, 2002 at 10:40 AM

I have two classes: CDog and CCat. I want CDog to contain a function pointer to a member function of CCat. I can't seem to get it to work quite right. Here is my code so far: CDog.h

  
class CDog
{
public:
	CDog();
	~CDog();
	static CCat* sm_pCat;

private:
	void LaunchFunction();
	void (CCat::*m_pRunFunction)();
};

CDog.cpp

  
#include "CDog.h"
#include "CCat.h"

CCat* CDog::sm_pCat = new CCat();

CDog::CDog() 
{
	m_pRunFunction = CCat::FunctionToPointTo;
}

CDog::~CDog() { /* ... */}

void CDog::LaunchFunction() { m_pRunFunction(); }		//error here

CCat.h

  
class CCat
{
public:
	CCat();
	~CCat();

	void FunctionToPointTo();
private:
};

I get an error saying that m_pRunFunction(); is not a valid function call. I'm pretty sure this code is almost right, but I can't seem to get it perfect. Any advice? -Mike [edited by - doctorsixstring on September 30, 2002 12:03:14 PM]

doctorsixstring

Author

doctorsixstring

Sep 30, 2002 at 1:52 PM

Does anyone have any ideas? I just checked out the other thread on this topic ("point to class member function."), but I couldn''t find anything. I tried some of the code people posted over there, but none of it wor that worked out. Any ideas at all would be appreciated.

-Mike

VolkerG

Sep 30, 2002 at 2:00 PM

If I am right, the syntax is

  
object.*m_pRunFunction ()
or
objectptr->*m_pRunFunction ()

where object is from the class CCat.

Auron

Sep 30, 2002 at 2:10 PM

Unless you declare the method you want to point to as static, you can only call a method by pointer if you make the call through an instance of that class.

So in your case, you''d have to do something like this:

void CDog::LaunchFunction (CCat *Cat) { Cat->(*m_pRunFunction)(); }

I think that''s it. The syntax may be a little off but the point is more what matters here:

You must use an instance to call a method pointer otherwise the method has no data to work from when called.

-Auron

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