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From: "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Newsgroups: comp.lang.c
Subject: Re: question about nullptr
Date: Tue, 9 Jul 2024 13:24:35 -0700
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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?