Messages in this thread | | | Date | Fri, 18 Jan 2008 12:47:01 +0100 | From | "Giacomo A. Catenazzi" <> | Subject | Re: Why is the kfree() argument const? |
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Jakob Oestergaard wrote: > On Thu, Jan 17, 2008 at 01:25:39PM -0800, Linus Torvalds wrote: > ... >> Why do you make that mistake, when it is PROVABLY NOT TRUE! >> >> Try this trivial program: >> >> int main(int argc, char **argv) >> { >> int i; >> const int *c; >> >> i = 5; >> c = &i; >> i = 10; >> return *c; >> } >> >> and realize that according to the C rules, if it returns anything but 10, >> the compiler is *buggy*. > > That's not how this works (as we obviously agree). > > Please consider a rewrite of your example, demonstrating the usefulness and > proper application of const pointers: > > extern foo(const int *); > > int main(int argc, char **argv) > { > int i; > > i = 5; > foo(&i); > return i; > } > > Now, if the program returns anything else than 5, it means someone cast away > const, which is generally considered a bad idea in most other software > projects, for this very reason. > > *That* is the purpose of const pointers.
"restrict" exists for this reason. const is only about lvalue.
You should draw a line, not to make C more complex!
Changing the name of variables in your example:
extern print_int(const int *);
int main(int argc, char **argv) { extern int errno;
errno = 0; print_int(&i); return errno; }
print_int() doesn't know that errno is also the argument. and this compilation unit doesn't know that print_int() will modify errno.
Ok, I changed int to extern int, but you see the point? Do you want complex rules about const, depending on context (extern, volatile,...) ?
ciao cate
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