Michael S. Tsirkin wrote: > On Mon, Sep 14, 2009 at 12:08:55PM -0400, Gregory Haskins wrote: >> Michael S. Tsirkin wrote: >>> On Fri, Sep 11, 2009 at 12:00:21PM -0400, Gregory Haskins wrote: >>>> FWIW: VBUS handles this situation via the "memctx" abstraction. IOW, >>>> the memory is not assumed to be a userspace address. Rather, it is a >>>> memctx-specific address, which can be userspace, or any other type >>>> (including hardware, dma-engine, etc). As long as the memctx knows how >>>> to translate it, it will work. >>> How would permissions be handled? >> Same as anything else, really. Read on for details. >> >>> it's easy to allow an app to pass in virtual addresses in its own address space. >> Agreed, and this is what I do. >> >> The guest always passes its own physical addresses (using things like >> __pa() in linux). This address passed is memctx specific, but generally >> would fall into the category of "virtual-addresses" from the hosts >> perspective. >> >> For a KVM/AlacrityVM guest example, the addresses are GPAs, accessed >> internally to the context via a gfn_to_hva conversion (you can see this >> occuring in the citation links I sent) >> >> For Ira's example, the addresses would represent a physical address on >> the PCI boards, and would follow any kind of relevant rules for >> converting a "GPA" to a host accessible address (even if indirectly, via >> a dma controller). > > So vbus can let an application "application" means KVM guest, or ppc board, right? > access either its own virtual memory or a physical memory on a PCI device. To reiterate from the last reply: the model is the "guest" owns the memory. The host is granted access to that memory by means of a memctx object, which must be admitted to the host kernel and accessed according to standard access-policy mechanisms. Generally the "application" or guest would never be accessing anything other than its own memory. > My question is, is any application > that's allowed to do the former also granted rights to do the later? If I understand your question, no. Can you elaborate? Kind Regards, -Greg
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