Re: [Qemu-devel] [PATCH] Permit zero-sized qemu_malloc() & friends

[malc]

On Mon, 7 Dec 2009, Markus Armbruster wrote:

> malc <address@hidden> writes:
> 
> > On Sun, 6 Dec 2009, Markus Armbruster wrote:
> >
> >> malc <address@hidden> writes:
> >> 
> >> > On Sun, 6 Dec 2009, Markus Armbruster wrote:
> >> >
> >> >> malc <address@hidden> 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
> 

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