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 <[email protected]>
Reviewed-by: Rik van Riel <[email protected]>
Cc: Linus Torvalds <[email protected]>
Cc: Suresh Siddha <[email protected]>
Cc: Andy Lutomirski <[email protected]>
Cc: Pekka Riikonen <[email protected]>
Link: http://lkml.kernel.org/r/[email protected]
Signed-off-by: Borislav Petkov <[email protected]>

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