How to devise function pointer type from function name?

[Nitin]

Hi,

Can I devise function pointer type from function name by using
template programming?
e.g.
I have function:
exertn “C” void xmlFreeDoc(xmlDocPtr cur);
I want something like following:
typeof(xmlFreeDoc) functionpointer = xmlFreeDoc;
i.e value to type conversion
I want pass this type to Holder which will call this function in dtor.

Currently I am using following code to do the same task:
extern "C" {
typedef void (*funcPtr)(void*) ;
}

children, 1), xmlFree);

Issues with above code:
I need to add every type of function in the extern “C” block.

Thanks,
Nitin.

 

[Juha Nieminen]

I have function:
exertn “C” void xmlFreeDoc(xmlDocPtr cur);
I want something like following:
typeof(xmlFreeDoc) functionpointer = xmlFreeDoc;

With the upcoming C++ standard you will be able to do it like this:

decltype(smlFreeDoc) functionpointer = xmlFreeDoc;

Or in this specific case, more easily:

auto functionpointer = xmlFreeDoc;

(decltype is still useful for some cases, where you need the type
itself, rather than just to create a variable.)