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c++ - need help void(s::*member func)()

reply 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 '(' 
}
Jan 25 2005
next sibling parent reply 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.
Jan 25 2005
parent 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.
Jan 26 2005
prev sibling parent 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.)
Jan 25 2005