Re: [PATCH 07/11] KVM: guest_memfd: extract __kvm_gmem_get_pfn()

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On Thu, Apr 04, 2024, Paolo Bonzini wrote:
> In preparation for adding a function that walks a set of pages
> provided by userspace and populates them in a guest_memfd,
> add a version of kvm_gmem_get_pfn() that has a "bool prepare"
> argument and passes it down to kvm_gmem_get_folio().
> Populating guest memory has to call repeatedly __kvm_gmem_get_pfn()
> on the same file, so make the new function take struct file*.
> Signed-off-by: Paolo Bonzini <pbonzini@xxxxxxxxxx>
> ---
>  virt/kvm/guest_memfd.c | 38 +++++++++++++++++++++++---------------
>  1 file changed, 23 insertions(+), 15 deletions(-)
> diff --git a/virt/kvm/guest_memfd.c b/virt/kvm/guest_memfd.c
> index 486748e65f36..a537a7e63ab5 100644
> --- a/virt/kvm/guest_memfd.c
> +++ b/virt/kvm/guest_memfd.c
> @@ -540,33 +540,29 @@ void kvm_gmem_unbind(struct kvm_memory_slot *slot)
>  	fput(file);
>  }
> -int kvm_gmem_get_pfn(struct kvm *kvm, struct kvm_memory_slot *slot,
> -		     gfn_t gfn, kvm_pfn_t *pfn, int *max_order)
> +static int __kvm_gmem_get_pfn(struct file *file, struct kvm_memory_slot *slot,
> +		       gfn_t gfn, kvm_pfn_t *pfn, int *max_order, bool prepare)

I genuinely don't know what it means to "prepare" a guest_memfd.  I see it becomes

	if (!prepare)
		fgp_flags |= FGP_CREAT_ONLY;

but I find the name "prepare" to be extremely unhelpful.

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