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

David Miller
2017-12-05T20:31:30Z

From: Geert Uytterhoeven <geert@linux-m68k.org>
Date: Tue, 5 Dec 2017 21:20:57 +0100

> 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.

Yeah, this is driving people nuts who wonder why pointers are aligned
all weird now.