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pointer to member functions

#include <stdio.h>
class a
{
public:
 void      funca();
};
class b
{
public:
void   funcb();
};

void a::funca() { printf("aaaaaaaaaaaaaa\n");};
void b::funcb() { printf("bbbbbbbbbbbbbb\n");};

class c:public a,public b
{
public:
      typedef void (c::*FUNC)();

      c(int aint){if (aint==0)
                  funcc=&c::funca;
                  else
                  funcc=&c::funcb;
                  
      };
      ~c(){};
public:
      void stubfunc(){(this->*funcc)();};//why need this-> ?
public:
   
      FUNC funcc;
};

void main()
{
      c ac(0),bc(1);
      (ac.*ac.funcc)();// why not be ac.*funcc()? more details will
                         // be appreciated.
      (ac.*bc.funcc)();// it's right! why?
      (bc.*ac.funcc)();
      (bc.*bc.funcc)();
      ac.stubfunc();
      bc.stubfunc();

}
0
anonexperts
Asked:
anonexperts
1 Solution
 
WynCommented:
hi,anonexperts:

one trival hint:

All non-static functions need a specific valid class instance to work with.

one friendly advice:

Here many experts on this field ,but you'd better give more points(at least 50pt).That will interst them to answer you and you will get more help.It's worthy.

Regards
W.Yinan

0
 
RONSLOWCommented:
>void stubfunc(){(this->*funcc)();};//why need this-> ?

->* is required to dereference the function pointer.

funcc() would not be enough.

You alwas need a LHS for the ->* (or .*) member-function pointer.  And in this case, it is a member of ourself (ie. this) that we want to call.

>(ac.*ac.funcc)();// why not be ac.*funcc()? more details will
>                         // be appreciated.

ac.*funcc would mean there is a local var called funcc that is a pointer-to-memeber-function.  In this case, the pointer to memeber function is a member of ac, so you need ac.*ac.func

>(ac.*bc.funcc)();// it's right! why?

yes .. this looks at bc to see which memeber function bc.funcc is pointing to, and then invokes that function for object ac.

0
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