NetNews Offline 2/NetNews Offline Volume 2.iso/news/comp/lang/c++-part2/14462

Text File | 1996-08-05 | 3.7 KB | 150 lines

Path: news.compuserve.com!newsmaster
From: Philippe Verdy <[e-mail]>
Newsgroups: comp.lang.c++,ualberta.cmput.201
Subject: Re: class/function pointer help!
Date: 31 Mar 1996 18:56:26 GMT
Organization: CompuServe Incorporated
Message-ID: <4jmkgq$[e-mail]>
NNTP-Posting-Host: hd80-162.compuserve.com

[e-mail] (Ryan Gallagher) s'Θcrit :
> Hi, I have a class that uses a function pointer. The problem Im having is
> that the pointer doesnt want to access the function when the function is a
> local function to the calling class. Here is what I mean....
>
> class expression
>        {
>         public:
>
>                 string S;
>         list L;
>         .
>         .
>         int   initlist();
>         .
>         int   isexpr(int);
>        };
>
>
> list, and string are both classes.
>
> char * string::POP( int (* separator)(int c) )
> {
>
>  int n=0,m=0,r,size=strlen(S);
>  char *A,*B,*C;
>
>   A=new char[size+1]; // allocate space for strings A & B
>    .
>    .
>
>
>     //  find the first occurence of a separator character
>     while(!(*separator)(A[n]) && A[n]) n++;
>
>     .
>     .
>     .
>   return A;
> }
>
> // heres the calling function
>
> int expression::initlist(void)
> {
>  char *temp;
>  float F;
>  char C;
>
>   L.init();
>
>   while((temp=S.POP( ::isexpr )))
>   {
>     F=C=0;
>     assert(convert(F,C,temp));
>     L.Add(L.Assign(F,C));
>     delete[] temp;
>   }
>
>  return 0;
> }
>
>
> this works when I define isexpr(int) globally, but I want to beable to
> call function expression::isexpr(int).....when I do this...
>
>   while((temp=S.POP( isexpr )))
>   {
>
>  the compiler say:" member function must be called or its address taken."
>
> how do I declare the member function as a function pointer?!
>
>
>
>
> thanks
> ------
> Please mail me back if possible.
>
This is not a problem of your compiler, but a C++ language
feature.
>
A member function is not a standard function.
So your isexpr() member function is not compatible with your
S.POP() formal argument.
This is because member functions have an hidden supplementary
argument which references the instance on which it applies.
The only way to manage with it would be to declare your
member function isexpr() as a static member function, so that
the instance reference is not passed. (So it will not be
available in the body of that function).

Declare S.POP() like this :
class string
{ ...
  static char * POP( int (* separator)(int c) );
}


The other way is to declare the type of the instance reference
(implicitly passed to your member function) within the
argument of S.POP(). Like this, the pointed function will have
access to its backgound instance.

Declare S.POP() like this :
class string
{ ...
  char * POP( int (expression::* separator)(int c) );
}

This explictly declares that the function pointed by
the "separator" argument, is a member of the "expression"
class, and so it has an implicit argument which is an
expression reference...

In summary, function pointers are not compatible to
non static member function pointers.
In addition, a method pointer from a given class is not compatible
with a method pointer from another class which is not
derived from the first one.

However, you can pass a pf method pointer like:
  int (X::*pf)(U, V);
as if it was in fact declared as :
  int (Y::*pf)(U, V);
provided X is derived from Y.

Its like if you had declared the function pointer:
  static int (*pf)(X& x, U u, V v);
and tried to pass it as an argument to a function which accepts
  static int (*pf)(Y& y, U u, V v);
and then use the x argument explicitly to access the members
of the actual y parameter.

The last declaration used is what you would use, if you have
to manage your pointers to several class types.

I hope this will help you...