• sergey (13/13) Jan 25 2005 Could anybody explain how to use type (fn_t) from the following example:
• Scott Michel (4/19) Jan 25 2005 Try "&a.fn" instead. After all, you're attempting to set a pointer to
• Don Clugston (32/51) Jan 26 2005 Even this isn't right. You can only write:
• Scott Michel (7/15) Jan 25 2005 Generally speaking, though, if you have to resort to pointers to member

sergey <sergey_member pathlink.com> writes:

Could anybody explain how to use type (fn_t) from the following example:

struct S {
int x;
void fn() { x++; }
};

typedef void (S::*fn_t)();

void main()
{
S a;
fn_t p;
p=NULL; // works  
p=a.fn; // doesn't work. error expecting '(' 
}

Scott Michel <scottm aero.org> writes:

sergey wrote:
 Could anybody explain how to use type (fn_t) from the following example:
 
 struct S {
 int x;
 void fn() { x++; }
 };
 
 typedef void (S::*fn_t)();
 
 void main()
 {
 S a;
 fn_t p;
 p=NULL; // works  
 p=a.fn; // doesn't work. error expecting '(' 

Try "&a.fn" instead. After all, you're attempting to set a pointer to 
point at the address of something. Otherwise, the parser is expecting a 
method invocation, thus, it expects a '(' to follow a.fn.

Don Clugston <Don_member pathlink.com> writes:

In article <ct66rn$2aso$1 digitaldaemon.com>, Scott Michel says...
sergey wrote:
 Could anybody explain how to use type (fn_t) from the following example:
 
 struct S {
 int x;
 void fn() { x++; }
 };
 
 typedef void (S::*fn_t)();
 
 void main()
 {
 S a;
 fn_t p;
 p=NULL; // works  
 p=a.fn; // doesn't work. error expecting '(' 

Try "&a.fn" instead. After all, you're attempting to set a pointer to 
point at the address of something. Otherwise, the parser is expecting a 
method invocation, thus, it expects a '(' to follow a.fn.

Even this isn't right. You can only write:
p = &S::fn;
You have to include the name of the class.
Note that "a" doesn't appear at all. You probably want it to call a.fn(), but
that requires a delegate, which you can't do legally in C++ (although Borland


Sergey, 
I didn't intend to plug my own article again, but I think you should read it...

http://www.codeproject.com/cpp/FastDelegate.asp

It was the most popular C++ article published on the CodeProject website in
2004, so you wouldn't be wasting your time. It gives a tutorial on member
function pointers, and provides a library that uses some tricks to allow you to
write code like:

#include "FastDelegate.h"

struct S {
int x;
void fn() { x++; }
};

typedef FastDelegate< void () > fn_t; // no parameters, returns void

void main()
{
S a;
fn_t p;
p.clear();
p.bind(&a, &S::fn); // or: p = MakeDelegate(&a, &S::fn);
p(); // calls a.fn()
}

As Scott says, you should seriously consider if you can use functors instead.
But functors won't let you do run-time polymorphism, so delegates can often be
very handy.

-Don.

Scott Michel <scottm aero.org> writes:

sergey wrote:
 Could anybody explain how to use type (fn_t) from the following example:
 
 struct S {
 int x;
 void fn() { x++; }
 };
 
 typedef void (S::*fn_t)();

Generally speaking, though, if you have to resort to pointers to member 
functions in C++, there is something hideously wrong with your design. 
Pointers to functions are much more useful if you're programming in C, 
but really wrong if you're programming in C++. In C++, you want to take 
advantage of functor structures (e.g., the one's STL likes that require 
you to implement an operator() method.)