__kernel_fpu_begin() does nothing if !__thread_has_fpu() && use_eager_fpu(),
perhaps it assumes that this case is simply impossible. This is certainly
not possible if in_interrupt() == T; interrupted_user_mode() should have
FPU, and interrupted_kernel_fpu_idle() should fail if !__thread_has_fpu().
However, even if use_eager_fpu() == T a task can do drop_fpu(), then switch
to another thread which becomes fpu_owner_task, then resume and call some
function which does kernel_fpu_begin(). Say, an exiting task does a lot of
things after exit_thread(), it is not safe to assume that it can't use FPU
in these paths.
Signed-off-by: Oleg Nesterov <o...@redhat.com>
---
arch/x86/kernel/i387.c | 5 +++--
1 files changed, 3 insertions(+), 2 deletions(-)
diff --git a/arch/x86/kernel/i387.c b/arch/x86/kernel/i387.c
index 81049ff..26f0e80 100644
--- a/arch/x86/kernel/i387.c
+++ b/arch/x86/kernel/i387.c
@@ -93,9 +93,10 @@ void __kernel_fpu_begin(void)
if (__thread_has_fpu(me)) {
__save_init_fpu(me);
- } else if (!use_eager_fpu()) {
+ } else {
this_cpu_write(fpu_owner_task, NULL);
- clts();
+ if (!use_eager_fpu())
+ clts();
}
}
EXPORT_SYMBOL(__kernel_fpu_begin);
--
1.5.5.1
--
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