Re: question about nullptr

On 7/9/2024 1:24 PM, Chris M. Thomasson wrote:

On 7/9/2024 3:14 AM, Ben Bacarisse wrote:

"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>
>

p = nullptr;

>
(p having been declared void *)
>

(0 == p == nullptr == NULL == 0) == true ?
>
Am I missing something here? If so, here is a preemptive: Shit!

>
You are missing that 0 == p == nullptr == NULL == 0 does not mean what
you want!  It means
>
  (((0 == p) == nullptr) == NULL) == 0
>
and that is a constraint violation.
>
Why?  Well 0 == p has value 1 and is of type int and equality
comparisons between int and nullptr_t values (of which there is only
one) are not permitted.  Catching this sort of thing is one of the
benefits of nullptr and its associated type nullptr_t.  It means that
while
>
#define nullptr ((void *)0)
>
can help to get C23 code to compile with a pre-C23 compiler, it might
result in some C23 constraint violations going undetected.
>
Anyway, that aside, I know what you meant.  To clarify, all of the
following have the value 1:
>
   0 == p
   p == nullptr
   nullptr == NULL
   NULL == 0
   0 == nullptr
   !p
>
Note that 0 == nullptr /is/ allowed even though 0 has type int.  That is
because 0 is also a null pointer constant, and the rules for == and !=
specifically allow it.
>

 It was a bit of pseudo code. Here is a program:
__________________________
#include <stdio.h>
#include <stdlib.h>
 int main()
{
     void* p = 0;
      if ((p == NULL) && (! p))
     {
         printf("Good!\n");
     }
      else
     {
         printf("Strange? Humm...\n");
     }
      return 0;
}
__________________________
 Good shall be printed even with the following condition right?
__________________________
if ((p == NULL) && (! p) && (p == nullptr))
{
    printf("Good!\n");
}
__________________________
 Any better Ben?

To be more precise, printf shall be called if p is 0, NULL or nullptr... They are all the same, in a sense, right?