int a[3], * restrict ip = &a[1]; *ip = 1; does a; invoke undefined behavior?
int a[3], * restrict ip = &a[1]; *ip = 1; does a; invoke undefined behavior?
No. There is no undefined behavior. Maybe there might be a compile error.
why do you think that?
Not sure what the question is? Clarify the question and and the code snippet.
Does it violate the guarantees of restrict? Is a; considered an access that conflicts with the modification of a[1] through a restrict qualified pointer?
Nope! as long as it is compiled in c99 mode.Originally Posted by robwhit
[Edit] If you don't know how to compile in c99 mode then replace restrict with __restrict__
Last edited by itCbitC; 06-02-2010 at 05:38 PM.
how do you know?
The way I see it, a restrict pointer means that only that pointer will be used to access the object(s) which it points to. In this case the object is the integer at a[1]. So if a[1] is accessed while ip is in scope, then behavior is undefined. Same thing goes for a[2] if *(ip+1) is accessed. You can access a[0] during the lifetime of ip, as long as you don't access *(ip-1).
gg
yes, I agree. but can you do 'a;'? note the difference between that and 'a[0];'. does 'a;' count as an access to a[0], a[1], and a[2]?
a is just an address (and the compiler can already track it's accesses). It would need to be dereferenced to be an "integer object" access.
gg
Last edited by Codeplug; 06-02-2010 at 09:59 PM. Reason: It
but before it's an address, it's an array object, correct? only then would it undergo array to pointer conversion.
let me put it another way...
struct S { int i; char c; };
S s = { 0 }, * restrict sp = &s; sp->i = 1;
would s; invoke undefined behavior? sure, it wouldn't need the values of any of its members in that statement, but what about this statement?
S s2 = s;
surely that would be an access of s, and violating sp's restrict qualifier, right? if "S s2 = s;" invokes undefined behavior, why wouldn't "s;"? why wouldn't 'a;' invoke undefined behavior if "s;" invokes undefined behavior?
Last edited by robwhit; 06-02-2010 at 10:20 PM. Reason: ;
>> would s; invoke undefined behavior?
I would say no. Because the compiler is not obligated to generate any real "access" (load or store) of "s".
The ip/sp examples aren't exactly the same:
The object that "sp" points to is "s".
The object that "ip" points to is "a[1]".
Access to the object "a[1]" is restricted only through "ip". So a, *a, *(a+2) are all ok and defined while ip is in scope.
Access to the object "s" is restricted only through "sp", but the expression "s;" doesn't constitute an access.
>> S s2 = s;
>> surely that would be an access of s
Yep.
gg
ah yes, I meant
struct S { int i; char c; };
S s = { 0 };
char * restrict cp = &s.c; *cp = 1;
your reasoning seems sane. thanks.
