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Em ter., 10 de nov. de 2020 às 10:43, Roberto Ierusalimschy <> escreveu:
> Ok, Makes sense to not use 0.
> In fact, I never use 0 with pointers, but I've seen a lot of code like this.
> Including IUP Lua.
> Why not memset with (int) NULL, wouldn't work?

The reality according to ISO C is subtle. 0 *is* the null pointer,
when it is a constant properly converted to a pointer: Pointers
  3 An integer constant _expression_ with the value 0, or such an
    _expression_ cast to type void *, is called a null pointer constant.(55)
    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.
  (55) The macro NULL is defined in <stddef.h> (and other headers) as
       a null pointer constant; see 7.17.

So, the "official" NULL in C is actually 0, NULL is only a macro.

The subtlelty is the cast: A null pointer is *a null pointer constant
converted to a pointer type*. A 0 converted to a float does not have
a zero representation; similarly, a constant 0 converted to a pointer
may not have a zero representation. And only a constant 0 converted
to a pointer is guaranteed to be NULL.

memset "fill[s] memory with a constant byte", not with a constant
pointer. So, the 0 (or NULL) given to it will never be a pointer,
and cannot properly represent the NULL pointer in an arbitrary
arquitecture, only in arquitectures where NULL is represented by
the same byte repeated multiple times. (In an 64-bit arquitecture
where the NULL is represented internally as 0xABABABABABABABAB,
«memset(m, 0xAB, len)» will initialize m with NULLs :-).

BTW, note that NULL is a "null pointer constant", not a "null
pointer" :-)
Ok, I think that it was very clear.
But C programmer shouldn't worry about filling memory with NULLs,
when C99 allow constructs like this:

/* printf example */
#include <stdio.h>

int main()
    char * names[32] = { NULL };
   printf ("name[30]=%p\n", names[30]);
   return 0;

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