From: Rik van Riel <r...@redhat.com>
With Oleg's patch "x86, fpu: don't abuse FPU in kernel threads if
use_eager_fpu()", kernel threads no longer have an FPU state even
on systems with use_eager_fpu()
That in turn means that a task may still have its FPU state
loaded in the FPU registers, if the task only got interrupted by
kernel threads from when it went to sleep, to when it woke up
again.
In that case, there is no need to restore the FPU state for
this task, since it is still in the registers.
The kernel can simply use the same logic to determine this as
is used for !use_eager_fpu() systems.
Signed-off-by: Rik van Riel <r...@redhat.com>
---
arch/x86/include/asm/fpu-internal.h | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/arch/x86/include/asm/fpu-internal.h
b/arch/x86/include/asm/fpu-internal.h
index 06af286593d7..723b74da0685 100644
--- a/arch/x86/include/asm/fpu-internal.h
+++ b/arch/x86/include/asm/fpu-internal.h
@@ -457,7 +457,7 @@ static inline fpu_switch_t switch_fpu_prepare(struct
task_struct *old, struct ta
task_disable_lazy_fpu_restore(old);
if (fpu.preload) {
new->thread.fpu_counter++;
- if (!use_eager_fpu() && fpu_lazy_restore(new, cpu))
+ if (fpu_lazy_restore(new, cpu))
fpu.preload = 0;
else
prefetch(new->thread.fpu.state);
--
1.9.3
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