Question

Interesting issue we came across here this week.

We are working in C on a Harvard Architecture embedded platform, which has 16-bit data addresses and 32-bit code addresses.

The issue occurs when you are working with function pointers. If you have code like

if (fp) fp();

or

if (fp != 0) fp();

everything is fine.

However if you have code like

if (fp != NULL) fp();

then, because NULL is defined as (void *) 0, the compiler (gcc in this case) a) does not warn and b) does a 16-bit comparison against your function pointer instead of a 32-bit comparison. Fine as long as your function pointer doesn’t happen to lie on a 64k boundary so all bottom 16 bits are 0.

At the moment we have large swathes of code which contain explicit checks against NULL. Most of them will be data pointers, but some of them will be function pointers. A quick grep for != NULL or == NULL revealed over 3000 results, to many to go through manually to check.

So, what we would like now would be either

1. a way to find all the cases where function pointers (but not data pointers) are compared (so we can instead have them compare against FP_NULL which we would define as a 32-bit 0), or

2. to redefine NULL in such a way that it would do the right thing.

3. (Or, I suppose, to
   update our gcc port to detect and
   correctly handle this case).

I can’t think of any approach that works for 1. The only approach I can think of for 2 is to redefine NULL as a 0 function pointer, which would be very wasteful for the vast majority of comparisons which are against data pointers. (A 32-bit compare is 4 instructions, a 16-bit compare is 1 instruction).

Any thoughts or suggestions?

Answer 1

It seems to me the easiest way is replace all occurrences of NULL by 0. This works for function pointer (as you say) and object pointers.

This is a variant of (2) Redefine NULL to plain 0.

But the fact that you cannot compare function pointers with NULL is a bug in your implementation. C99 states that comparison of the null pointer constant is possible with both object and function pointers, and that NULL should expand to this constant.

Small addition from the C-FAQ question 5.8:

Q: Is NULL valid for pointers to functions?
A: Yes (but see question 4.13)

Mixing function pointers with (void *) 0

(A reply to R..’s comment). I believe using function pointers and (void *) 0 together is well-defined. In my reasoning I will refer to sections of the C99 draft 1256, but will not quote large parts to keep it readable. It should also be applicable to C89.

• 6.3.2.3 (3) defines the integer constant expression 0 and such an expressions cast to (void *) as null pointer constant. And: “If a null pointer constant is converted to a
  pointer type, the resulting pointer, called a null pointer, is guaranteed to compare unequal
  to a pointer to any object or function.”
• 6.8.9 defines the == and != operands for (among other) a pointer operand and a null pointer constant. For these: “If one operand is a pointer and the other is a
  null pointer constant, the null pointer constant is converted to the type of the pointer.”

Conclusion: In fp == (void *) 0, the null pointer constant is converted to the type of fp. This null pointer can be compared to fp and is guaranteed to be unequal to fp if it points to a function. Assignment (=) has a similar clause, so fp = (void *) 0; is also well-defined C.