If you mean you want the program image to be shared in main memory, then that's already done automatically for read-only parts of the image.
If you mean you want the program image to be shared on...
Type: Posts; User: robwhit
If you mean you want the program image to be shared in main memory, then that's already done automatically for read-only parts of the image.
If you mean you want the program image to be shared on...
Elysia, what you're talking about reminds me of what Mach was trying to do. Basically, it was (is) a microkernel, with everything possible moved into its own process (task), separate from the...
Ebay.
You have 6 %fs in your format strings. Don't you want 1?
int fclose(FILE *fp2); this is a function declaration. A declaration tells the compiler of something (contrasting to a definition which...
I don't know.
You didn't #include the header file that declared TransparentBlt.
Those are turned around. !first is equivalent to first == NULL, and current->next in a condition (like in an if statement or loop condition) is equivalent to current->next != NULL.
Ok, seems I'm wrong somewhat. in N1256 (a post-C99 draft), there's footnote 82 which says:So it appears to be valid, as long as the accessed value isn't a trap representation in the type you're using...
whups sorry. knee-jerk reaction.
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.
are you sure it's not just starting with str and a lowercase letter?
better yet,
double array[10]; // for example
size_t size = sizeof(array);sizeof evaluates to a value of type size_t, which is an unsigned integer (int != integer, int is a kind of integer) type.
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 }, *...
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]?
how do you know?
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?
Sure it's not cheating, but it is undefined behavior in ISO standard C. You could use compiler extensions like GCC has to get the behavior you want, though.
why do you think that?
int a[3], * restrict ip = &a[1]; *ip = 1; does a; invoke undefined behavior?
Hi Dave. You've set a great example here and I keep thinking I should act more like you. Sorry you're sick, I hope you get better.
I guess we miss each other's points then.
You're saying that a part of your body seems to resemble a Turing machine in some way, therefore all life is a Turing machine? Even disregarding the fact that you completely miss what "life" is,...
No, you are confused. A Turing machine involves an strip of tape of infinite length, symbols encoded on that tape, and something that changes those symbols. That has nothing to do with life.
Hmm. Well, you're arguing minor semantics of word definitions, which I am not.
If you think that the phrase "Tomorrow is green!" means that "tomorrow" and "green" mean the same thing, you don't...
Because it's like saying green is the same thing as tomorrow. They're completely different.