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.
Signed-off-by: Ram Pai <redacted>
---
arch/powerpc/mm/pkeys.c | 14 +++++++++-----
1 files changed, 9 insertions(+), 5 deletions(-)
@@ -19,6 +19,7 @@u64pkey_amr_mask;/* Bits in AMR not to be touched */u64pkey_iamr_mask;/* Bits in AMR not to be touched */u64pkey_uamor_mask;/* Bits in UMOR not to be touched */+intexecute_only_key=2;#define AMR_BITS_PER_PKEY 2#define AMR_RD_BIT 0x1UL
@@ -122,8 +122,12 @@ int pkey_initialize(void)#elseos_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);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));for(i=(pkeys_total-os_reserved);i<pkeys_total;i++)pkey_uamor_mask&=~(0x3ul<<pkeyshift(i));
Key allocation and deallocation has the side effect of programming the
UAMOR/AMR/IAMR registers. This is wrong, since its the responsibility of
the application and not that of the kernel, to modify the permission on
the key.
Do not modify the pkey registers at key allocation/deallocation.
This patch also fixes a bug where a sys_pkey_free() resets the UAMOR
bits of the key, thus making its permissions unmodifiable from user
space. Latter if the same key gets reallocated from a different thread
this thread will no longer be able to change the permissions on the key.
Problem noticed/reported by Michael Ellermen while running
selftests/core-pkeys
Signed-off-by: Ram Pai <redacted>
---
arch/powerpc/include/asm/pkeys.h | 11 -----------
arch/powerpc/mm/pkeys.c | 27 ---------------------------
2 files changed, 0 insertions(+), 38 deletions(-)
@@ -224,33 +224,6 @@ static inline void init_iamr(int pkey, u8 init_bits)write_iamr(old_iamr|new_iamr_bits);}-staticvoidpkey_status_change(intpkey,boolenable)-{-u64old_uamor;--/* Reset the AMR and IAMR bits for this key */-init_amr(pkey,0x0);-init_iamr(pkey,0x0);--/* Enable/disable key */-old_uamor=read_uamor();-if(enable)-old_uamor|=(0x3ul<<pkeyshift(pkey));-else-old_uamor&=~(0x3ul<<pkeyshift(pkey));-write_uamor(old_uamor);-}--void__arch_activate_pkey(intpkey)-{-pkey_status_change(pkey,true);-}--void__arch_deactivate_pkey(intpkey)-{-pkey_status_change(pkey,false);-}-/**SettheaccessrightsinAMRIAMRandUAMORregistersfor@pkeytothat*specifiedin@init_val.
Only when the key is allocated, its permission are enabled.
Signed-off-by: Ram Pai <redacted>
---
tools/testing/selftests/powerpc/ptrace/core-pkey.c | 4 ++++
1 files changed, 4 insertions(+), 0 deletions(-)
The test case assumes execute-permissions of unallocated keys are
enabled by default.
Signed-off-by: Ram Pai <redacted>
---
.../testing/selftests/powerpc/ptrace/ptrace-pkey.c | 5 +++++
1 files changed, 5 insertions(+), 0 deletions(-)
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.
@@ -19,6 +19,7 @@u64pkey_amr_mask;/* Bits in AMR not to be touched */u64pkey_iamr_mask;/* Bits in AMR not to be touched */u64pkey_uamor_mask;/* Bits in UMOR not to be touched */+intexecute_only_key=2;#define AMR_BITS_PER_PKEY 2#define AMR_RD_BIT 0x1UL
@@ -122,8 +122,12 @@ int pkey_initialize(void)#elseos_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.
quoted hunk
initial_allocation_mask = reserved_allocation_mask | (0x1 << PKEY_0);
/* register mask is in BE format */
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
From: Gabriel Paubert <hidden> Date: 2018-06-29 06:08:20
On Thu, Jun 28, 2018 at 11:56:34PM -0300, Thiago Jung Bauermann wrote:
Hello,
Ram Pai [off-list ref] writes:
quoted
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.
@@ -19,6 +19,7 @@u64pkey_amr_mask;/* Bits in AMR not to be touched */u64pkey_iamr_mask;/* Bits in AMR not to be touched */u64pkey_uamor_mask;/* Bits in UMOR not to be touched */+intexecute_only_key=2;#define AMR_BITS_PER_PKEY 2#define AMR_RD_BIT 0x1UL
@@ -122,8 +122,12 @@ int pkey_initialize(void)#elseos_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?
Not in general. It probably always works on Power because of the definition
of the machine instruction for shifts with variable amount (consider the
shift amount unsigned and take it modulo twice the width of the operand),
but for example it fails on x86 (1<<-1 gives 0x80000000).
If so, a comment pointing this out would make this less confusing.
Unless I miss something, this code is run once at boot, so its
performance is irrelevant.
In this case simply rewrite it as:
reserved_allocation_mask = 0x1 << 1;
if ( (pkeys_total - os_reserved) <= execute_only_key) {
execute_only_key = -1;
} else {
reserved_allocation_mask = (0x1 << 1) | (0x1 << execute_only_key);
}
Caveat, I have assumed that the code will either:
- only run once,
- pkeys_total and os_reserved are int, not unsigned
quoted
initial_allocation_mask = reserved_allocation_mask | (0x1 << PKEY_0);
/* register mask is in BE format */
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?
Same answer: very likely on Power, not portable.
Gabriel
On Thu, Jun 28, 2018 at 11:56:34PM -0300, Thiago Jung Bauermann wrote:
quoted
Hello,
Ram Pai [off-list ref] writes:
quoted
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.
@@ -19,6 +19,7 @@u64pkey_amr_mask;/* Bits in AMR not to be touched */u64pkey_iamr_mask;/* Bits in AMR not to be touched */u64pkey_uamor_mask;/* Bits in UMOR not to be touched */+intexecute_only_key=2;#define AMR_BITS_PER_PKEY 2#define AMR_RD_BIT 0x1UL
@@ -122,8 +122,12 @@ int pkey_initialize(void)#elseos_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?
Not in general. It probably always works on Power because of the definition
of the machine instruction for shifts with variable amount (consider the
shift amount unsigned and take it modulo twice the width of the operand),
Ok, thanks for confirming.
but for example it fails on x86 (1<<-1 gives 0x80000000).
Strange, this works on my laptop with an Intel(R) Core(TM) i5-7300U CPU:
$ cat blah.c
#include <stdio.h>
int main(int argc, char *argv[])
{
printf("1 << -1 = %llx\n", (unsigned long long) 1 << -1);
return 0;
}
$ make blah
cc blah.c -o blah
blah.c: In function 'main':
blah.c:5:52: warning: left shift count is negative [-Wshift-count-negative]
printf("1 << -1 = %llx\n", (unsigned long long) 1 << -1);
^~
$ ./blah
1 << -1 = 0
Even if I change the cast and printf format to int, the result is the
same. Or am I doing it wrong?
quoted
If so, a comment pointing this out would make this less confusing.
Unless I miss something, this code is run once at boot, so its
performance is irrelevant.
In this case simply rewrite it as:
reserved_allocation_mask = 0x1 << 1;
if ( (pkeys_total - os_reserved) <= execute_only_key) {
execute_only_key = -1;
} else {
reserved_allocation_mask = (0x1 << 1) | (0x1 << execute_only_key);
}
I agree it's clearer and more robust code (except that the first
line should be inside the if block).
Caveat, I have assumed that the code will either:
- only run once,
- pkeys_total and os_reserved are int, not unsigned
Both of the above are true.
--
Thiago Jung Bauermann
IBM Linux Technology Center
On Fri, Jun 29, 2018 at 09:58:37PM -0300, Thiago Jung Bauermann wrote:
Gabriel Paubert [off-list ref] writes:
quoted
On Thu, Jun 28, 2018 at 11:56:34PM -0300, Thiago Jung Bauermann wrote:
quoted
Hello,
Ram Pai [off-list ref] writes:
quoted
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.
@@ -19,6 +19,7 @@u64pkey_amr_mask;/* Bits in AMR not to be touched */u64pkey_iamr_mask;/* Bits in AMR not to be touched */u64pkey_uamor_mask;/* Bits in UMOR not to be touched */+intexecute_only_key=2;#define AMR_BITS_PER_PKEY 2#define AMR_RD_BIT 0x1UL
@@ -122,8 +122,12 @@ int pkey_initialize(void)#elseos_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?
Not in general. It probably always works on Power because of the definition
of the machine instruction for shifts with variable amount (consider the
shift amount unsigned and take it modulo twice the width of the operand),
Ok, thanks for confirming.
quoted
but for example it fails on x86 (1<<-1 gives 0x80000000).
Strange, this works on my laptop with an Intel(R) Core(TM) i5-7300U CPU:
$ cat blah.c
#include <stdio.h>
int main(int argc, char *argv[])
{
return 0;
}
$ make blah
cc blah.c -o blah
blah.c: In function 'main':
blah.c:5:52: warning: left shift count is negative [-Wshift-count-negative]
printf("1 << -1 = %llx\n", (unsigned long long) 1 << -1);
^~
$ ./blah
1 << -1 = 0
My intel box does the same. It makes it zero. So does my
powerpc box. Mathematically, (1 << -1) is nothing but 2^-1,
which is 1/2, which is 0.5, and when rounded it has to be 0.
However, yes, GCC defines it to be 'undefined'. gcc compiler does
warn 'left shift count is negative'. Will have to fix it.
Thanks for catching this.
Even if I change the cast and printf format to int, the result is the
same. Or am I doing it wrong?
quoted
quoted
If so, a comment pointing this out would make this less confusing.
Unless I miss something, this code is run once at boot, so its
performance is irrelevant.
In this case simply rewrite it as:
reserved_allocation_mask = 0x1 << 1;
if ( (pkeys_total - os_reserved) <= execute_only_key) {
execute_only_key = -1;
} else {
reserved_allocation_mask = (0x1 << 1) | (0x1 << execute_only_key);
}
I tried hard not to sprikle if-then-else in the code. Makes it less
elegant. But then, correctness is more important than
elegance.
I agree it's clearer and more robust code (except that the first
line should be inside the if block).
quoted
Caveat, I have assumed that the code will either:
- only run once,
- pkeys_total and os_reserved are int, not unsigned
From: Gabriel Paubert <hidden> Date: 2018-06-30 16:57:13
On Fri, Jun 29, 2018 at 09:58:37PM -0300, Thiago Jung Bauermann wrote:
Gabriel Paubert [off-list ref] writes:
quoted
On Thu, Jun 28, 2018 at 11:56:34PM -0300, Thiago Jung Bauermann wrote:
quoted
Hello,
Ram Pai [off-list ref] writes:
quoted
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.
@@ -19,6 +19,7 @@u64pkey_amr_mask;/* Bits in AMR not to be touched */u64pkey_iamr_mask;/* Bits in AMR not to be touched */u64pkey_uamor_mask;/* Bits in UMOR not to be touched */+intexecute_only_key=2;#define AMR_BITS_PER_PKEY 2#define AMR_RD_BIT 0x1UL
@@ -122,8 +122,12 @@ int pkey_initialize(void)#elseos_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?
Not in general. It probably always works on Power because of the definition
of the machine instruction for shifts with variable amount (consider the
shift amount unsigned and take it modulo twice the width of the operand),
Ok, thanks for confirming.
quoted
but for example it fails on x86 (1<<-1 gives 0x80000000).
Strange, this works on my laptop with an Intel(R) Core(TM) i5-7300U CPU:
$ cat blah.c
#include <stdio.h>
int main(int argc, char *argv[])
{
printf("1 << -1 = %llx\n", (unsigned long long) 1 << -1);
return 0;
}
$ make blah
cc blah.c -o blah
blah.c: In function 'main':
blah.c:5:52: warning: left shift count is negative [-Wshift-count-negative]
printf("1 << -1 = %llx\n", (unsigned long long) 1 << -1);
^~
$ ./blah
1 << -1 = 0
Even if I change the cast and printf format to int, the result is the
same. Or am I doing it wrong?
Try something more dynamic, 1 << -1 is evaluated at compile time by gcc,
and when you get a warning, the compile time expression evaluation does
not always give the same result as the run time machine instructions.
To test this I wrote (yes, the error checking is approximate, it would
be better to use strtol):
/***************************************************************************/
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char **argv)
{
int val, amount, valid;
if (argc != 3) {
printf("Needs 2 arguments!\n");
return EXIT_FAILURE;
}
valid = sscanf(argv[1], "%d", &val);
if (valid == 1) {
valid = sscanf(argv[2], "%d", &amount);
}
if (valid == 1) {
printf("%d shifted by %d is %d\n", val, amount, val<<amount);
return EXIT_SUCCESS;
} else {
printf("Both arguments must be integers!\n");
return EXIT_FAILURE;
}
}
/***************************************************************************/
Compile it, and then run it with 1 and -1 as parameters. The result is
not the same on PPC and x86.
quoted
quoted
If so, a comment pointing this out would make this less confusing.
Unless I miss something, this code is run once at boot, so its
performance is irrelevant.
In this case simply rewrite it as:
reserved_allocation_mask = 0x1 << 1;
if ( (pkeys_total - os_reserved) <= execute_only_key) {
execute_only_key = -1;
} else {
reserved_allocation_mask = (0x1 << 1) | (0x1 << execute_only_key);
}
I agree it's clearer and more robust code (except that the first
line should be inside the if block).
Indeed, sorry for this.
Gabriel
quoted
Caveat, I have assumed that the code will either:
- only run once,
- pkeys_total and os_reserved are int, not unsigned
Both of the above are true.
--
Thiago Jung Bauermann
IBM Linux Technology Center
Key allocation and deallocation has the side effect of programming the
UAMOR/AMR/IAMR registers. This is wrong, since its the responsibility of
the application and not that of the kernel, to modify the permission on
the key.
Do not modify the pkey registers at key allocation/deallocation.
This patch also fixes a bug where a sys_pkey_free() resets the UAMOR
bits of the key, thus making its permissions unmodifiable from user
space. Latter if the same key gets reallocated from a different thread
this thread will no longer be able to change the permissions on the key.
Problem noticed/reported by Michael Ellermen while running
selftests/core-pkeys
Signed-off-by: Ram Pai <redacted>
---
arch/powerpc/include/asm/pkeys.h | 11 -----------
arch/powerpc/mm/pkeys.c | 27 ---------------------------
2 files changed, 0 insertions(+), 38 deletions(-)
LGTM.
Reviewed-by: Thiago Jung Bauermann <redacted>
--
Thiago Jung Bauermann
IBM Linux Technology Center
Only when the key is allocated, its permission are enabled.
Signed-off-by: Ram Pai <redacted>
---
tools/testing/selftests/powerpc/ptrace/core-pkey.c | 4 ++++
1 files changed, 4 insertions(+), 0 deletions(-)
The test case assumes execute-permissions of unallocated keys are
enabled by default.
Signed-off-by: Ram Pai <redacted>
---
.../testing/selftests/powerpc/ptrace/ptrace-pkey.c | 5 +++++
1 files changed, 5 insertions(+), 0 deletions(-)