Pointer to void != pointer to function

Question

Hello C experts,
this is a simple question about SunStudio's understanding of function pointers (it might have been asked before):
Why is the following C code giving me a warning on all Sun compilers:
# cat funcptr.c
int main() {
    void (*func)() = (void*)0;
    return 0;
# cc -c funcptr.c
"funcptr.c", line 2: warning: assignment type mismatch:
     pointer to function() returning void "=" pointer to voidIs the warning emitted by Sun compiler a bug (over-cautions). Other compiler's don't warn here.
Of course, I could use 0 or NULL as rvalue but I'm just wondering...
Jochen

Answer 1

Thanks everybody for your opinions.
@Stephen:
I wasn't aware that data pointers and function pointers may differ in any aspect. In fact, I only know data models that talk about 'pointer' width in general, like LP64 (64-bit wide long- and pointertypes), ILP32 (32-bit wide int-, long- and pointer-types), .. I'm especially surprised to hear that casting between data/function pointers will result in undefined behavior.
Anyway, to properly 'fix' the code above, I must cast the zero to a function pointer (with matching signature) instead of a void pointer. This compiles w/o a warning:
    void (*func)() = (void (*)())0;Thanks for clarification.
Jochen