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From: dan@BBN-PROPHET.ARPA
Newsgroups: net.lang.c
Subject: Re:  Array of pointers to functions
Message-ID: <2450@brl-smoke.ARPA>
Date: Mon, 7-Apr-86 16:29:29 EST
Article-I.D.: brl-smok.2450
Posted: Mon Apr  7 16:29:29 1986
Date-Received: Thu, 10-Apr-86 00:43:18 EST
Sender: news@brl-smoke.ARPA
Lines: 46

Re array of pointers to functions: the way I think of it is that C
declarations must be read "from the inside out".  Start with the
identifier itself, and work your way out from the tightest-binding
declaration operator to the loosest.  The first declaration operator you
hit gives you the fundamental type of the object.  So, given a declaration
containing

	... (*x) ...

No matter what is surrounding the "(*x)", x is a pointer.  Always.
Never an array, function, etc., though what it points to may involve
these.  On the other hand, given

	... (x[5]) ...

X is definitely an array of 5 (somethings).  Maybe pointers, ints, etc.
So, to construct an array of pointers, you say

	(x[])

to get the array, and add the '*' next:

	(*(x[]))

Now it's an array of pointers.  What do they point to?  Functions:

	( (*(x[])) () )

Now all you need to do is specify what the functions return:

     int ( (*(x[])) () )

And you're done.  However, you've left behind a pretty confusing
declaration.  C typedefs are the perfect way to make things like this easy
to understand:

    typedef int FUNC();
    typedef FUNC * FUNC_PTR;
    FUNC_PTR x[];

Now it's quite clear that x is an array of function pointers.  I usually
use some variation on this when I have to construct such arrays; it's a
lot clearer, even to people like me who have been using C for years.
Hope this helps.

	Dan Franklin