[PATCH v5.5 23/30] KVM: Resolve memslot ID via a hash table instead of via a static array
Maciej S. Szmigiero
maciej.szmigiero at oracle.com
Thu Nov 11 15:51:47 PST 2021
On 04.11.2021 01:25, Sean Christopherson wrote:
> From: Maciej S. Szmigiero <maciej.szmigiero at oracle.com>
>
> Memslot ID to the corresponding memslot mappings are currently kept as
> indices in static id_to_index array.
> The size of this array depends on the maximum allowed memslot count
> (regardless of the number of memslots actually in use).
>
> This has become especially problematic recently, when memslot count cap was
> removed, so the maximum count is now full 32k memslots - the maximum
> allowed by the current KVM API.
>
> Keeping these IDs in a hash table (instead of an array) avoids this
> problem.
>
> Resolving a memslot ID to the actual memslot (instead of its index) will
> also enable transitioning away from an array-based implementation of the
> whole memslots structure in a later commit.
>
> Signed-off-by: Maciej S. Szmigiero <maciej.szmigiero at oracle.com>
> Co-developed-by: Sean Christopherson <seanjc at google.com>
> Signed-off-by: Sean Christopherson <seanjc at google.com>
> ---
> include/linux/kvm_host.h | 16 +++----
> virt/kvm/kvm_main.c | 96 +++++++++++++++++++++++++++++++---------
> 2 files changed, 84 insertions(+), 28 deletions(-)
>
(..)
> @@ -1259,17 +1257,49 @@ static int kvm_alloc_dirty_bitmap(struct kvm_memory_slot *memslot)
> return 0;
> }
>
> +static void kvm_replace_memslot(struct kvm_memslots *slots,
> + struct kvm_memory_slot *old,
> + struct kvm_memory_slot *new)
> +{
> + /*
> + * Remove the old memslot from the hash list, copying the node data
> + * would corrupt the list.
> + */
> + if (old) {
> + hash_del(&old->id_node);
> +
> + if (!new)
> + return;
> + }
> +
> + /* Copy the source *data*, not the pointer, to the destination. */
> + if (old)
> + *new = *old;
This way of writing it (that, is re-checking whether "old" is not-NULL)
suggests that it could have been set to NULL inside the previous block
(since the last check), which isn't true.
Thanks,
Maciej
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