Re: NULL pointer arithmetic issues

[Johnny Billquist <bqt%softjar.se@localhost>]

On 2020-02-25 02:12, Mouse wrote:
> > Oh.  And I actually do not believe it has to be a constant.
>
> You are correct; it does not need to be a simple constant.
>
> > The text says "integer constant expression with the value 0, or such
> > an expression..."
>
> Yes.  (void *)(1-1) is a valid null pointer constant.  So, on an
> all-ASCII system, is (0x1f+(3*5)-'.').  But, in the presence of
>
> int one(void) { return(1); }
>
> then (one()-one()) is not - it is an integer expression with value
> zero, but it is not an integer _constant_ expression.  It's entirely
> possible that (int *)(one()-one()) will produce a different pointer
> from (int *)(1-1) - the latter is a null pointer; the former might or
> might not be, depending on the implementation.

As you say, it's an integer expression. And I read that "or" part as 
just an expression, which this is. So I believe it is a valid way to 
creating something that can be converted to a NULL pointer.

Also:

if (expression) statement; shall execute statement if expression not 
equals 0, according to the standard.

So, where does that leave this code:

char *p;

.
.

if (p) foo();

p is not an integer. How do you compare it to 0?

  Johnny

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Johnny Billquist                  || "I'm on a bus
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