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.comSigned-off-by: Borislav Petkov <bp@suse.de>

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);