returning a void (*)(void)

[Sergio]

If I have a private member:

void (*func)(void);

how can i declare a 'get' function that returns it? I tryed:

void (*)()GetFunc()
{
return func;
}

but looks like that's not the way...

thanks

[Bob Hairgrove]

On 5 Jan 2005 03:57:37 -0800, "Sergio" <se*******@gmail.com> wrote:

If I have a private member:

void (*func)(void);

how can i declare a 'get' function that returns it? I tryed:

void (*)()GetFunc()
{
return func;
}

but looks like that's not the way...

thanks

It's easier with a typedef:

typedef void (* func_t)();

struct C {
func_t func;
func_t getFunc() { return func; }
};

--
Bob Hairgrove
No**********@Home.com

[Dietmar Kuehl]

Sergio wrote:

If I have a private member:

void (*func)(void);
Note that this signature matches a member only if the member
is static: otherwise the type is more something like this

| void (C::*func)(void);

where 'C' is the class containing the member.
how can i declare a 'get' function that returns it? I tryed:

void (*)()GetFunc()

| void (*GetFunc())()

is the syntax for a void function returning void.
--
<mailto:di***********@yahoo.com> <http://www.dietmar-kuehl.de/>
<http://www.contendix.com> - Software Development & Consulting

[owl ling]

well, yoiu should writen as the following code.
typedef void (*FUNC)(void);

void Func(void)
{
cout << "func" << endl;
}
FUNC GetFunc()
{
FUNC fp = Func;
return fp;
}

[msalters]

Sergio wrote:

If I have a private member:

void (*func)(void);

how can i declare a 'get' function that returns it? I tryed:

void (*)()GetFunc()
{
return func;
}

but looks like that's not the way...

Don't try it this way, use a typedef.

typedef void(*fun_void_void)();
fun_void_void func;
fun_void_void GetFunc() { return func; }
HTH,
Michiel Salters

[Old Wolf]

msalters wrote:

Sergio wrote:

If I have a private member:

void (*func)(void);
how can i declare a 'get' function that returns it? I tryed:

Don't try it this way, use a typedef.

typedef void(*fun_void_void)();
fun_void_void func;
fun_void_void GetFunc() { return func; }

Another option is to typedef the function type (rather than the
pointer-to-function). I only mention this because it hadn't
occurred to me that it was possible until recently, and I prefer
to avoid pointer typedefs if I can:

| typedef void (fun_void_void) ();
| fun_void_void func; // this declares a function
| fun_void_void *ptr_func = func;
| fun_void_void * GetFunc() { return ptr_func; }

or

| fun_void_void * GetFunc() { return func; }

because the name of a function is converted to a pointer to
that function, in a value context.

[Jonathan Turkanis]

Sergio wrote:

If I have a private member:

void (*func)(void);

how can i declare a 'get' function that returns it? I tryed:

On more way:

#include <boost/mpl/identity.hpp>

using namespace boost::mpl;

identity<void (*)(void)>::type get();