Re: [PATCH V11 3/5] printk: hash addresses printed with %p

Geert Uytterhoeven
2017-12-05T20:21:01Z

Hi Tobin,

On Wed, Nov 29, 2017 at 3:05 AM, Tobin C. Harding <me@tobin.cc> wrote:
> Currently there exist approximately 14 000 places in the kernel where
> addresses are being printed using an unadorned %p. This potentially
> leaks sensitive information regarding the Kernel layout in memory. Many
> of these calls are stale, instead of fixing every call lets hash the
> address by default before printing. This will of course break some
> users, forcing code printing needed addresses to be updated.
>
> Code that _really_ needs the address will soon be able to use the new
> printk specifier %px to print the address.

> --- a/lib/vsprintf.c
> +++ b/lib/vsprintf.c

> +/* Maps a pointer to a 32 bit unique identifier. */
> +static char *ptr_to_id(char *buf, char *end, void *ptr, struct printf_spec spec)
> +{
> + unsigned long hashval;
> + const int default_width = 2 * sizeof(ptr);
> +
> + if (unlikely(!have_filled_random_ptr_key)) {
> + spec.field_width = default_width;
> + /* string length must be less than default_width */
> + return string(buf, end, "(ptrval)", spec);
> + }
> +
> +#ifdef CONFIG_64BIT
> + hashval = (unsigned long)siphash_1u64((u64)ptr, &ptr_key);
> + /*
> + * Mask off the first 32 bits, this makes explicit that we have
> + * modified the address (and 32 bits is plenty for a unique ID).
> + */
> + hashval = hashval & 0xffffffff;
> +#else
> + hashval = (unsigned long)siphash_1u32((u32)ptr, &ptr_key);
> +#endif

Would it make sense to keep the 3 lowest bits of the address?

Currently printed pointers no longer have any correlation with the actual
alignment in memory of the object, which is a typical cause of a class of bugs.

Gr{oetje,eeting}s,

Geert

--
Geert Uytterhoeven -- There's lots of Linux beyond ia32 -- geert@linux-m68k.org

In personal conversations with technical people, I call myself a hacker. But
when I'm talking to journalists I just say "programmer" or something like that.
-- Linus Torvalds