How do I cast to "pointer to function returning int" ?

Tim McDaniel mcdaniel at amara.uucp
Fri Apr 13 10:16:48 AEST 1990

CMH117 at (Charles Hannum) writes:
   Casting a void function to an int function is extremely non-portable.

Not quite.  ANSI C, and many existing compilers, requires that
function pointers all be the same size.  Any pointer to a function can
be cast into any other pointer-to-function type and back again,
producing the same pointer value.  However, the value can't be called
while it is cast to a different type.  For example:

   typedef anytype   (*PF_ANY  )();
   typedef othertype (*PF_OTHER)();
    * PF_ANY is the type "pointer to function () returning anytype".
    * PF_OTHER is the type "pointer to function () returning othertype".
    * Assume that anytype and othertype are different types.
   PF_ANY   f1, f2;
   PF_OTHER g;
   g  = (PF_OTHER) f1;	  /* legal under ANSI C */
   f2 = (PF_ANY) g;	  /* legal under ANSI C */
   if (f1 != f2) {
      /* then this compiler is not ANSI C compliant. */
   (*f2)();		  /* exact same result as "(*f1)();" */
   (*g)();		  /* illegal under ANSI C */

And, by the way,
   f2 = (PF_ANY) (void *) f1;
is illegal in ANSI C.  "void *" is the universal pointer only for DATA
objects, not functions.  You can convert any FUNCTION pointer to any
other FUNCTION pointer type.

To rephrase, then:
   Casting a pointer to void function to a pointer to int function
   type, and calling that result, is extremely non-portable.

Tim McDaniel
Applied Dynamics International, Ann Arbor, MI
Internet: mcdaniel%amara.uucp at
UUCP: {uunet,sharkey}!amara!mcdaniel

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