Path: utzoo!utgpu!jarvis.csri.toronto.edu!mailrus!ames!pasteur!ucbvax!tut.cis.ohio-state.edu!rutgers!att!ulysses!andante!alice!ark
From: ark@alice.UUCP (Andrew Koenig)
Newsgroups: comp.lang.c
Subject: Re: function casting
Message-ID: <9266@alice.UUCP>
Date: 30 Apr 89 15:49:23 GMT
References: <12481@umn-cs.CS.UMN.EDU>
Organization: AT&T Bell Laboratories, Liberty Corner NJ
Lines: 57

In article <12481@umn-cs.CS.UMN.EDU>, clark@umn-cs.CS.UMN.EDU (Robert P. Clark) writes:

> How do I cast something to a pointer to a function that returns
> an integer? One of my (failed) attempts has been


> main()
> {
>   int  (*f)();
>   char *foo();


>     f = ((*int)())foo();  /*  foo returns char*, but I know this is  */

> }

Best would be to make foo return an appropriate pointer ane
avoid type cheating:

	int (*foo())();
	int (*f)();

	f = foo();

The declaration of foo here says that if you call foo, dereference the
result, and call that, you get an int.  The definition of foo would
look something like this:

	int (*foo())()
	{
		/* ... */
	}

Don't be too astonished if your compiler balks at this -- but
it is indeed valid C.

Next question: how to declare a variable to contain a pointer to
a function that returns an integer?  If f is such a variable,
then to get an integer from it, you dereference it (to get a
function from the function pointer) and then call the function.
As your example shows, such a declaration looks like this:

	int (*f)();

Finally, how do you cast a value to the type of f, assuming
you still want to?  A cast looks a lot like the declaration
with parentheses around it, and with the variable removed
that was declared.  So int(*f)() becomes (int(*)()) and the
assignment looks something like this:

	f = (int(*)()) foo();

Of course, the cast is unnecessary if you get the type of
foo right to begin with.  [C-T&P, p. 13]
-- 
				--Andrew Koenig
				  ark@europa.att.com