On 2019-01-25 16:18:40 [+0100], Borislav Petkov wrote: > Reviewed-by: Borislav Petkov <bp@xxxxxxx> thanks. > Should we do this microoptimization in addition, to save us the > activation when the kernel thread here: > > taskA -> kernel thread -> taskA > > doesn't call kernel_fpu_begin() and thus fpu_fpregs_owner_ctx remains > the same? This might work now but at the end of the series this case will be handled. The switch taskA -> kernel thread will save taskA's registers. The switch kernel thread -> taskA will only set TF flag to restore FPU registers on the return to userland. The load happens only the ctx pointer is different. > It would be a bit more correct as it won't invoke the > trace_x86_fpu_regs_activated() TP in case the FPU context is the same. The trace point is not wrong. As of now the same context will be loaded again. Sebastian
