Passing function addresses as parameters (Calling convention problems)

[Mike]

I want to pass the address of a function to a timeout type class so that
it can be used as a callback function.

The function I want to use as a callback is defined like...

void rf04::rc_proc(void)
{
}

The call to the the function that uses this functions' address is like...

rctimer=new Timeout(rf04::rc_proc);

But the compiler complains with ...

h:\work\telemetry\rf04.cpp(81) : error C2664: 'Timeout::Timeout' :
cannot convert parameter 1 from 'void (void)' to 'void (__cdecl *)(void)'

I tried defining my callback function as

_CRTIMP void __cdecl rf04::rc_proc(void)

but then the compiler complained that I was not allowed to use dllimport
functions in this way.

Linux and g++ complains in a different way...

g++ -Wall -g -O -c rf04.cpp
rf04.cpp: In method `rf04::rf04 (char *)':
rf04.cpp:82: assuming & on `rf04::rc_proc'
rf04.cpp:82: no matching function for call to `Timeout::Timeout (void
(rf04::*) ())'
timeout.h:19: candidates are: Timeout::Timeout (void (*) (), unsigned
int = 1000)
timeout.h:29: Timeout::Timeout (const Timeout &)
make: *** [rf04.o] Error 1

I'm at bit of a loss to understand what is going on. I never really
understood #pragma and stuff like that, but I suspect it's siomething to
do with that.

If my call back function was a global function and not part of a class
then the code compiles and runs OK.

Can anyone give me some clues on this.

Thanks.

[Victor Bazarov]

Mike wrote:

I want to pass the address of a function to a timeout type class so that
it can be used as a callback function.

The function I want to use as a callback is defined like...

void rf04::rc_proc(void)
{
}

The call to the the function that uses this functions' address is like...

rctimer=new Timeout(rf04::rc_proc);

But the compiler complains [...]

This is covered in the FAQ. Please see the section on pointers to
members.

V