On 01/10/2020 09:12 PM, Thomas Gleixner wrote:
Christophe Leroy <christophe.le...@c-s.fr> writes:
diff --git a/lib/vdso/gettimeofday.c b/lib/vdso/gettimeofday.c
index 17b4cff6e5f0..5a17a9d2e6cd 100644
--- a/lib/vdso/gettimeofday.c
+++ b/lib/vdso/gettimeofday.c
@@ -144,7 +144,7 @@ __cvdso_gettimeofday(const struct vdso_data *vd, struct
__kernel_old_timeval *tv
static __maybe_unused __kernel_old_time_t
__cvdso_time(const struct vdso_data *vd, __kernel_old_time_t *time)
{
- __kernel_old_time_t t =
READ_ONCE(vd[CS_HRES_COARSE].basetime[CLOCK_REALTIME].sec);
+ __kernel_old_time_t t = vd[CS_HRES_COARSE].basetime[CLOCK_REALTIME].sec;
if (time)
*time = t;
Allows the compiler to load twice, i.e. the returned value might be different
from the
stored value. So no.
With READ_ONCE() the 64 bits are being read:
00000ac8 <__c_kernel_time>:
ac8: 2c 03 00 00 cmpwi r3,0
acc: 81 44 00 20 lwz r10,32(r4)
ad0: 81 64 00 24 lwz r11,36(r4)
ad4: 41 82 00 08 beq adc <__c_kernel_time+0x14>
ad8: 91 63 00 00 stw r11,0(r3)
adc: 7d 63 5b 78 mr r3,r11
ae0: 4e 80 00 20 blr
Without the READ_ONCE() only 32 bits are read. That's the most optimal.
00000ac8 <__c_kernel_time>:
ac8: 7c 69 1b 79 mr. r9,r3
acc: 80 64 00 24 lwz r3,36(r4)
ad0: 4d 82 00 20 beqlr
ad4: 90 69 00 00 stw r3,0(r9)
ad8: 4e 80 00 20 blr
Without READ_ONCE() but with a barrier() after the read, we should get
the same result but GCC (GCC 8.1) does less good:
00000ac8 <__c_kernel_time>:
ac8: 81 24 00 24 lwz r9,36(r4)
acc: 2f 83 00 00 cmpwi cr7,r3,0
ad0: 41 9e 00 08 beq cr7,ad8 <__c_kernel_time+0x10>
ad4: 91 23 00 00 stw r9,0(r3)
ad8: 7d 23 4b 78 mr r3,r9
adc: 4e 80 00 20 blr
Assuming both part of the 64 bits data will fall into a single
cacheline, the second read is in the noise.
So agreed to drop this change.
Christophe
