>> int one(void) { return(1); }
>> then (one()-one()) is not [a null pointer constant]
> 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.
C99 words it as
[#3] An integer constant expression with the value 0, or
such an expression cast to type void *, is called a null
pointer constant.
That little word "such" is important. As I read it, that means a null
pointer constant is "[a]n integer constant expression with the value 0,
or an integer constant expression with the value 0 cast to type void *"
(not just "an integer expression with the value 0 cast to...").
> if (expression) statement; shall execute statement if expression not
> equals 0, according to the standard.
Yes:
[#2] In both forms, the first substatement is executed if
the expression compares unequal to 0. In the else form, the
second substatement is executed if the expression compares
equal to 0. If the first substatement is reached via a
label, the second substatement is not executed.
> So, where does that leave this code:
> char *p;
> if (p) foo();
> p is not an integer. How do you compare it to 0?
The same way you do in
if (p != 0) foo();
How else?
I would admittedly prefer slightly more verbose wording, saying
explicitly that the 0 to which the control expression is compared is,
when applicable, a null pointer constant. (Comparison of a pointer
with a null pointer constant is specifically permitted for == and != -
see 6.5.9.)
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