Apple

template <typename R>

struct FTypes {

typedef R (*rt)(...);

typedef R (*rt_void)();

};

template <typename R>

typename FTypes<R>::rt convert_ptr(typename FTypes<R>::rt_void src) {

return reinterpret_cast<typename FTypes<R>::rt>(src);

}

is it correct? if so, I'm just curious, why do you need it?

thanks

On Dec 18, 2009, at 9:15 AM, Dan Caugherty wrote:

Thank you very much!   (Nice catch!)

Cheers,
-- Dan C.

On Fri, Dec 18, 2009 at 4:49 AM, Jens Miltner <email@hidden> wrote:

Am 18.12.2009 um 04:17 schrieb Dan Caugherty:

Hey all --

This is really a C/C++ question, and it's giving me fits, since it's not obvious why there's a syntax error.

Now, I believe I have a legitimate reason for doing what I'm doing, so just trust me. Yes, the code below looks dangerous, but I have some function pointers in legacy code that I must deal with, so I'm trying to minimize the danger in this particular case with some modicum of type checking based on return type.

I'd like to write a quick function template to change a function pointer from type " return_type (*)(void) " to "return_type (*)(...)". I thought this would work:

template <typename RetT> // zero-parameter case
RetT (*convert_ptr(...)) ( RetT (*src)(void) ) { // can't use variadic templates yet.. sigh....
   typedef RetT (*rt)(...); //wouldn't mind getting rid of this line somehow

  return reinterpret_cast<rt>(src); //ERROR: "src" was not declared in this scope.
}

Why is gcc tripping on the last line?

I believe the template function declaration needs to be :

RetT (*convert_ptr ( RetT (*src)(void) ))(...) { // can't use variadic templates yet.. sigh....

</jum>

Dmitry Markman

References:
	>Awful function pointer conversion function template (From: Dan Caugherty <email@hidden>)
	>Re: Awful function pointer conversion function template (From: Jens Miltner <email@hidden>)
	>Re: Awful function pointer conversion function template (From: Dan Caugherty <email@hidden>)