On 01/19/2015 01:51 PM, Oleg Nesterov wrote:
__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>
Reviewed-by: Rik van Riel <r...@redhat.com>
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