Re: [PATCH v2 02/17] x86/hyperv: detect if Linux is the root partition

[Wei Liu <wei.liu@kernel.org>]

On Thu, Nov 12, 2020 at 04:16:30PM +0100, Vitaly Kuznetsov wrote:
[...]
> >  /*
> >   * Virtual processor will never share a physical core with another virtual
> >   * processor, except for virtual processors that are reported as sibling SMT
> > diff --git a/arch/x86/include/asm/mshyperv.h b/arch/x86/include/asm/mshyperv.h
> > index ffc289992d1b..ac2b0d110f03 100644
> > --- a/arch/x86/include/asm/mshyperv.h
> > +++ b/arch/x86/include/asm/mshyperv.h
> > @@ -237,6 +237,8 @@ int hyperv_fill_flush_guest_mapping_list(
> >  		struct hv_guest_mapping_flush_list *flush,
> >  		u64 start_gfn, u64 end_gfn);
> >  
> > +extern bool hv_root_partition;
> 
> Eventually this is not going to be an x86 only thing I believe?

I hope so. :-)

> 
> > +
> >  #ifdef CONFIG_X86_64
> >  void hv_apic_init(void);
> >  void __init hv_init_spinlocks(void);
> > diff --git a/arch/x86/kernel/cpu/mshyperv.c b/arch/x86/kernel/cpu/mshyperv.c
> > index 05ef1f4550cb..f7633e1e4c82 100644
> > --- a/arch/x86/kernel/cpu/mshyperv.c
> > +++ b/arch/x86/kernel/cpu/mshyperv.c
> > @@ -237,6 +237,22 @@ static void __init ms_hyperv_init_platform(void)
> >  	pr_debug("Hyper-V: max %u virtual processors, %u logical processors\n",
> >  		 ms_hyperv.max_vp_index, ms_hyperv.max_lp_index);
> >  
> > +	/*
> > +	 * Check CPU management privilege.
> > +	 *
> > +	 * To mirror what Windows does we should extract CPU management
> > +	 * features and use the ReservedIdentityBit to detect if Linux is the
> > +	 * root partition. But that requires negotiating CPU management
> > +	 * interface (a process to be finalized).
> > +	 *
> > +	 * For now, use the privilege flag as the indicator for running as
> > +	 * root.
> > +	 */
> > +	if (cpuid_ebx(HYPERV_CPUID_FEATURES) & HV_CPU_MANAGEMENT) {
> 
> We may want to cache cpuid_ebx(HYPERV_CPUID_FEATURES) somewhere but we
> already had a discussion regading naming for these caches and decided to
> wait until TLFS for ARM is out so we don't need to rename again.

Exactly.

Wei.