On Mon, 7 Dec 2009, Markus Armbruster wrote: > malc <av1...@comtv.ru> writes: > > > On Sun, 6 Dec 2009, Markus Armbruster wrote: > > > >> malc <av1...@comtv.ru> writes: > >> > >> > On Sun, 6 Dec 2009, Markus Armbruster wrote: > >> > > >> >> malc <av1...@comtv.ru> writes: > >> >> > >> > > >> > [..snip..] > >> > > >> >> > >> >> read(fd, malloc(0), 0) is just fine, because read() doesn't touch the > >> >> buffer when the size is zero. > >> >> > >> > > >> > [..snip..] > >> > > >> > Yet under linux the address is checked even for zero case. > >> > >> Any value you can obtain from malloc() passes that check. > >> > >> Why does the fact that you can construct pointers that don't pass this > >> check matter for our discussion of malloc()? > >> > >> >> > I don't know what a "valid pointer" in this context represents. > >> >> > >> >> I can talk standardese, if you prefer :) > >> >> > >> >> malloc() either returns either a null pointer or a pointer to the > >> >> allocated space. In either case, you must not dereference the pointer. > >> >> > >> >> OpenBSD chooses to return a pointer to the allocated space. It chooses > >> >> to catch common ways to dereference the pointer. > >> >> > >> >> Your "p = (void *)-1" is neither a null pointer nor can it point to > >> >> allocated space on your particular system. Hence, it cannot be a value > >> >> of malloc() for any argument, and therefore what read() does with it on > >> >> that particular system doesn't matter. > >> >> > >> > > >> > Here, i believe, you are inventing artificial restrictions on how > >> > malloc behaves, i don't see anything that prevents the implementor > >> > from setting aside a range of addresses with 31st bit set as an > >> > indicator of "zero" allocations, and then happily giving it to the > >> > user of malloc and consumming it in free. > >> > >> Misunderstanding? Such behavior is indeed permissible, and I can't see > >> where I restricted it away. An implementation that behaves as you > >> describe returns "pointer to allocated space". That the pointer has > >> some funny bit set doesn't matter. That it can't be dereferenced is > >> just fine. > >>
Here you agree that it's permissible. > >> I'm not sure what your point is. If it is that malloc(0) can return a > >> value that cannot be passed to a zero-sized read(), then I fear you have > >> not made your point. > > > > One more attempt to make it clearer. If you agree that this behaviour > > is permissible then the game is lost as things stand now under Linux, > > since replacing [1]: > > > > void *p = (void *) -1 > > with: > > void *p = (void *) 0x80000000 > > > > or anything else with said bit set will yield EFAULT. Consequently the > > code you cited as a well behaving malloc(0) call site will bomb. > > > > [1] Under 32bit Linux that is, with the usual split. > > You can't just pull pointers out of your ear and expect stuff to work. And here you don't. Which renders whole discussion rather pointless. > > malloc() is free to return a pointer to allocated space that is set up > in a way that catches access beyond the allocated size. OpenBSD does > that for size zero; it allocates one byte then, from pages that are used > only for zero-sized allocations, and takes care to disable access to > these pages with mprotect(..., PROT_NONE)[*]. Since read(..., 0) does > not access beyond the allocated size, it still works just fine. > > If you replace glibc's malloc() to get OpenBSD-like behavior, you can't > just make up some pointer to a memory area you believe to be unused, you > have to do it right, like OpenBSD does. > > > [*] Check out omalloc_make_chunks() at > http://www.openbsd.org/cgi-bin/cvsweb/src/lib/libc/stdlib/malloc.c?rev=1.121;content-type=text%2Fplain > -- mailto:av1...@comtv.ru
