[PATCH 1/2] powerpc/pkeys: preallocate execute_only key only if the key is available.
Thiago Jung Bauermann
bauerman at linux.ibm.com
Fri Jun 29 12:56:34 AEST 2018
Hello,
Ram Pai <linuxram at us.ibm.com> writes:
> Key 2 is preallocated and reserved for execute-only key. In rare
> cases if key-2 is unavailable, mprotect(PROT_EXEC) will behave
> incorrectly. NOTE: mprotect(PROT_EXEC) uses execute-only key.
>
> Ensure key 2 is available for preallocation before reserving it for
> execute_only purpose. Problem noticed by Michael Ellermen.
Since "powerpc/pkeys: Preallocate execute-only key" isn't upstream yet,
this patch could be squashed into it.
> Signed-off-by: Ram Pai <linuxram at us.ibm.com>
> ---
> arch/powerpc/mm/pkeys.c | 14 +++++++++-----
> 1 files changed, 9 insertions(+), 5 deletions(-)
>
> diff --git a/arch/powerpc/mm/pkeys.c b/arch/powerpc/mm/pkeys.c
> index cec990c..0b03914 100644
> --- a/arch/powerpc/mm/pkeys.c
> +++ b/arch/powerpc/mm/pkeys.c
> @@ -19,6 +19,7 @@
> u64 pkey_amr_mask; /* Bits in AMR not to be touched */
> u64 pkey_iamr_mask; /* Bits in AMR not to be touched */
> u64 pkey_uamor_mask; /* Bits in UMOR not to be touched */
> +int execute_only_key = 2;
>
> #define AMR_BITS_PER_PKEY 2
> #define AMR_RD_BIT 0x1UL
> @@ -26,7 +27,6 @@
> #define IAMR_EX_BIT 0x1UL
> #define PKEY_REG_BITS (sizeof(u64)*8)
> #define pkeyshift(pkey) (PKEY_REG_BITS - ((pkey+1) * AMR_BITS_PER_PKEY))
> -#define EXECUTE_ONLY_KEY 2
>
> static void scan_pkey_feature(void)
> {
> @@ -122,8 +122,12 @@ int pkey_initialize(void)
> #else
> os_reserved = 0;
> #endif
> +
> + if ((pkeys_total - os_reserved) <= execute_only_key)
> + execute_only_key = -1;
> +
> /* Bits are in LE format. */
> - reserved_allocation_mask = (0x1 << 1) | (0x1 << EXECUTE_ONLY_KEY);
> + reserved_allocation_mask = (0x1 << 1) | (0x1 << execute_only_key);
My understanding is that left-shifting by a negative amount is undefined
behavior in C. A quick test tells me that at least on the couple of
machines I tested, 1 < -1 = 0. Does GCC guarantee that behavior? If so,
a comment pointing this out would make this less confusing.
> initial_allocation_mask = reserved_allocation_mask | (0x1 << PKEY_0);
>
> /* register mask is in BE format */
> @@ -132,11 +136,11 @@ int pkey_initialize(void)
>
> pkey_iamr_mask = ~0x0ul;
> pkey_iamr_mask &= ~(0x3ul << pkeyshift(PKEY_0));
> - pkey_iamr_mask &= ~(0x3ul << pkeyshift(EXECUTE_ONLY_KEY));
> + pkey_iamr_mask &= ~(0x3ul << pkeyshift(execute_only_key));
>
> pkey_uamor_mask = ~0x0ul;
> pkey_uamor_mask &= ~(0x3ul << pkeyshift(PKEY_0));
> - pkey_uamor_mask &= ~(0x3ul << pkeyshift(EXECUTE_ONLY_KEY));
> + pkey_uamor_mask &= ~(0x3ul << pkeyshift(execute_only_key));
Here the behaviour is undefined in C as well, given that pkeyshift(-1) =
64, which is the total number of bits in the left operand. Does GCC
guarantee that the result will be 0 here as well?
--
Thiago Jung Bauermann
IBM Linux Technology Center
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