Only if pointers are 8 bits.

More specifically, "NUL byte" is the character that has value 0, whereas "null pointer" is the pointer that has value 0.

OK, I'll try to stop being the pedant. I'll let the real pedants poke holes in what I said...

> More specifically, "NUL byte" is the character that has value 0, whereas "null pointer" is the pointer that has value 0.

Where is this enforced? You can assign between notional types without a problem. Whatever the context, providing 0 will have the same effect no matter how you labeled it.

Enforced? Not, as I think you're pointing out, by prohibiting conversion. That is, if

  char c = '\0';
  void* p = (void*)0;
  int a = (int)c;
  int b = (int)p;

then

  a == b

resolves to true. In that sense, they are the same - which I believe is your point. In that point, you are completely correct.

The difference is that I can't do

  *c

In that sense, they are not the same - not in the sense of numberic value, but in the sense of type. Also, c is 8 bits, and p (at least these days) is 32 or 64. (I pity anyone who ever had to work in an environment where p was 8 bits!) So they are the same numerically, but they are different both in type and in memory footprint.

a == p will also resolve to true. So will c == p. a and b don't add anything to your example; they're just in there to make c and p look like they're more different than they are.

    $ cat zero.c
    #include "stdio.h"

    int main(int argc, char* argv[]) {
     char nul = 0;
     void* null = 0;

     if( nul == null ) {
      printf("compared char to pointer; they are the same\n");
     } else {
      printf("found a difference between char and pointer\n");
     }
     return 0;
    }

    $ gcc -o zero zero.c
    zero.c: In function ‘main’:
    zero.c:7:10: warning: comparison between pointer and integer
      if( nul == null ) {
              ^~
    $ ./zero
    compared char to pointer; they are the same
    $

You get a warning, but not an error, for making the comparison. By contrast, assigning the integer zero to a void* isn't even a warning -- it's just a natural thing to do. There isn't another way to set a pointer to NULL. There is another way to set a character to 0, the '\0' syntax, but that's not a warning either.

C will think nothing of adding '!' to 'P' and getting 'q'. That's not strange because addition is a pretty normal thing to do with integers. You're right that a char variable should only occupy 8 bits of memory, but that's an implementation artifact, not a theory of what the value '\0' means. That value is unambiguously the integer zero with infinite precision. The reason it only occupies 8 bits is that you can't let it have infinite bits.

Hmm. Probably true. What about c == p? I don't recall whether both will promote to numeric, or whether it's a type error.

For that matter, what does (uncasted) p = c do? How about c = p?

[Edit: You updated while I was writing; my first question you already answered. Warning, not error.]

    $ cat pointers.c 
    #include "stdio.h"

    int main(int argc, char* argv[]) {
     char Z = 'Z';
     char q = 'q';
     void* null = 0;

     printf("Z is \\x%02x\n", Z);
     printf("But if it were a pointer, it would be %08x\n", Z);
     printf("Watch this:\n\n");

     null = Z;

     /* %p to print a pointer value */
     printf("Our void* is now: %p\n", null);

     q = null;

     printf("And q is: %c\n", q);

     return 0;
    }
    $ gcc -o pointers pointers.c
    pointers.c: In function ‘main’:
    pointers.c:12:7: warning: assignment makes pointer from integer without a cast [-Wint-conversion]
      null = Z;
           ^
    pointers.c:17:4: warning: assignment makes integer from pointer without a cast [-Wint-conversion]
      q = null;
        ^
    $ ./pointers
    Z is \x5a
    But if it were a pointer, it would be 0000005a
    Watch this:

    Our void* is now:     0x5a
    And q is: Z
    $

The assignments are warnings. They work just like you'd expect them to work.

Notice all the different printf flags? This is why you need them.

The literal 0 in a pointer context will be converted to a NULL pointer, which can be a non-zero bit pattern (there are some systems where the actual NULL pointer isn't all zeros). Going through a variable might not do what you think. So this:

    char *p = 0;

is fine, but

    intptr_t a = 0;
    char *p = (char *)a;

might not do what you expect (set p to the NULL pointer).

That’s not accurate. NUL byte is a byte with an all-zeros bit pattern, where a null pointer is a special pointer value (and thus pointer-sized) that can be coerced from the integer literal 0, but not in general from an arbitrary integer with value zero, and what’s more, a null pointer value is not guaranteed to have an all-zero bit pattern!