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x86/fpu: __kernel_fpu_begin() should clear fpu_owner_task even if use_eager_fpu()
__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 <oleg@redhat.com> Reviewed-by: Rik van Riel <riel@redhat.com> Cc: Linus Torvalds <torvalds@linux-foundation.org> Cc: Suresh Siddha <sbsiddha@gmail.com> Cc: Andy Lutomirski <luto@amacapital.net> Cc: Pekka Riikonen <priikone@iki.fi> Link: http://lkml.kernel.org/r/20150119185132.GB16427@redhat.com Signed-off-by: Borislav Petkov <bp@suse.de>
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@ -93,9 +93,10 @@ void __kernel_fpu_begin(void)
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if (__thread_has_fpu(me)) {
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__save_init_fpu(me);
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} else if (!use_eager_fpu()) {
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} else {
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this_cpu_write(fpu_owner_task, NULL);
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clts();
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if (!use_eager_fpu())
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clts();
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}
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}
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EXPORT_SYMBOL(__kernel_fpu_begin);
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