and a pointer to struct S:
    struct S * pS;

and if   pS= &S;

then I can call function f in two ways:

     (*(pS->f))();
or
     pS.f();
My question is why does this last method work?  I though the dot
operator only worked with the name of a structure, not a pointer
to it?
thanks for any insight,
James

--

   If I have a structure:
     struct S_type
         {
         ....
         (void)(*f)(void);
         ....
         } S;

   and a pointer to struct S:
       struct S * pS;

   and if   pS= &S;

   then I can call function f in two ways:

        (*(pS->f))();
   or
        pS.f();
   My question is why does this last method work?  I though the dot
   operator only worked with the name of a structure, not a pointer
   to it?

It shouldn't work AFAIK.  However, the following method should:
   pS->f();
You can call a function that you have a pointer to without needing to
dereference it.
--
"It takes a certain amount of shamelessness
 to be a monomaniac billionaire dwarf."
--Jon Katz <URL:http://slashdot.org/articles/99/03/17/1634238.shtml>
--

        pS->f();

It is perhaps not well known that the operator () applies to pointers
to functions and not to functions. If g is a function then in the call
g() 'g' would be implicitly converted to a pointer to g.

Thus - formally - you are dereferencing pS->f to yield a "function"
and then this function will be implicitly re-converted to a pointer to
the function (which f already is) because the () operator is applied
to it. Your code is correct, but it is an unnecessary NOP.

        (*pS).f()

would be OK, though, and is identical to  pS->f()

Regards
Horst

--