From: Vik Heyndrickx <Vik DOT Heyndrickx AT rug DOT ac DOT be>
Newsgroups: comp.lang.c,comp.os.msdos.djgpp
Subject: Re: Casting void pointers
Date: Tue, 23 Jun 1998 09:06:38 +0200
Organization: University of Ghent, Belgium
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Todd Rowan wrote:
> 
> How do you explicity cast a void pointer to a function pointer? Or do you
> even need to cast void pointers manually?

To be able to cast a function pointer to a void pointer is a DJGPP
extension to the C language. It is not standard available in all C
implementations or on all platforms. Please consider that when using it!

  void fn(double,double); // example: fn's prototype
  void *vp;

  vp = (void*)fn; // On older C implementations (pre-ANSI) you had to
write (void*)&fn

Now the answer to the question (note that each brace is required):

  fn's type is ((void*)(double,double))

So casting vp to a function pointer of type ((void*)(double,double))
gives:
  
  ((void*)(double,double))vp

When you want to call this function with a number of arguments, you have
to write e.g.:

  (((void*)(double,double))vp)(3.0,-2.0);

Note that the excessive amount of braces is required.
Note also that I haven't tested any of this, but it should work.

-- 
 \ Vik /-_-_-_-_-_-_/
  \___/ Heyndrickx /
   \ /-_-_-_-_-_-_/  Knight in the Order of the Unsigned Types