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SubjectRe: [PATCH] Use NULL instead of integer 0 in security/selinux/


On Sat, 10 Jul 2004, Alexandre Oliva wrote:
>
> > Why should they be?
>
> Err... Because the conditional expression is implicitly compared with
> 0 [6.8.4.1]/#2. If 0 is not to be used explicitly in pointer
> contexts, why should it be ok to use it implicitly?

Exactly BECAUSE it's not an explicitly WRONG type of 0.

"0" is a number token. It is totally illogical from a conceptual
standpoint to use it as a pointer. It makes no sense from any syntactic
standpoint, and it's very much an ugly special case because K&R didn't
want to add a keyword for NULL.

But the fact is, even early on, exactly _because_ "0" is illogical as a
pointer, K&R added a

#define NULL 0

to make it make _syntactic_ sense to use "NULL" as a pointer, even though
the language lacked the specific keyword. So from a _syntactic_
standpoint, NULL is a pointer, even if from an implementation standpoint
NULL ended up being this totally illogical integer 0.

In contrast, there is nothing syntactically strange about comparing a
non-boolean (even though Pascal and other languages make it illegal). So
here again, the "compare against 0" is an _implementation_ issue, not a
conceptual syntactic confusion.

What I object to in using "0" as a pointer is that it changes the meaning
of the token "0" depending on semantic information that may not even be
very local. In contrast

if (ptr)
..

has no such confusion.

> [6.5.3.3]/#5 defines the result of the logical negation operator
> based on the result of comparing the expression with 0.

And you're totally confusing the "this is defined to be equivalent" as an
implementation standpoint with "it's the same thing".

For example, the code

5[ptr]

is _defined_ to be exactly the same as

ptr[5]

since they both really mean

*(ptr + 5)

and nothing else. HOWEVER, despite the fact that the C language _defines_
that they are exactly equivalent, I claim that anybody who writes "5[ptr]"
is an ass.

For that same reason, your argument is totally irrelevant. Yes,

if (ptr)

is _defined_ to be exactly the same as

if (ptr != 0)

but I claim that anybody who writes the latter is an ass, because it makes
no syntactic sense. The same way "5[ptr]" doesn't make syntactic sense,
even though the compiler will silently accept it.

Got it?

Linus
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