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From: | Alexander Graf |
Subject: | [Qemu-devel] Re: [PATCH 2/3] Assume PPC64 host on PPC32 KVM |
Date: | Fri, 24 Jul 2009 13:23:30 +0200 |
On 24.07.2009, at 13:17, Jan Kiszka wrote:
Alexander Graf wrote:On 24.07.2009, at 12:59, Jan Kiszka wrote:Alexander Graf wrote:When talking to the kernel about dirty maps, we need to find out whichbits were actually set. This is done by set_bit and test_bit like functiontality which uses the "long" variable type. Now, with PPC32 userspace and PPC64 kernel space (which is pretty common),we can't interpret the bits properly anymore, because we think long is32 bits wide.So for PPC dirty bitmap analysis, let's just assume we're always running on a PPC64 host. Currently there is no dirty bitmap implementation forPPC32 / PPCEMB anyways. Unbreaks dirty logging on PPC. Signed-off-by: Alexander Graf <address@hidden> --- kvm-all.c | 6 ++++++ 1 files changed, 6 insertions(+), 0 deletions(-) diff --git a/kvm-all.c b/kvm-all.c index 824bb4c..bfaa623 100644 --- a/kvm-all.c +++ b/kvm-all.c @@ -357,7 +357,13 @@ int kvm_physical_sync_dirty_bitmap(target_phys_addr_t start_addr, for (phys_addr = mem->start_addr, addr = mem->phys_offset; phys_addr < mem->start_addr + mem->memory_size;phys_addr += TARGET_PAGE_SIZE, addr += TARGET_PAGE_SIZE) {+#ifdef HOST_PPC+ /* Big endian keeps us from having different long sizesin user and + * kernel space, so assume we're always on ppc64. */ + uint64_t *bitmap = (uint64_t *)d.dirty_bitmap; +#else unsigned long *bitmap = (unsigned long *)d.dirty_bitmap; +#endif unsigned nr = (phys_addr - mem->start_addr) >> TARGET_PAGE_BITS; unsigned word = nr / (sizeof(*bitmap) * 8); unsigned bit = nr % (sizeof(*bitmap) * 8);This rather screams for a generic fix. Current code assumes sizeof(unsigned long) == 8. That should already break on 32-bit x86hosts. So either do (sizeof(*bitmap) * sizeof(unsigned long)) or switchto uint64_t - but for ALL hosts.I don't see where that would break. The kernel treats the array as ulong*, userspace treats it as ulong* and set_bit in kernel doesbitmap[word] |= (1 << bit). So as long as userspace long and kernel longare the same, it works.In fact - it should even work out with little endian and different ulongsizes. It just breaks on BE.Err, yes, forget it. But let's help me understanding the actual problem: Do you have different ulong sizes in your scenario? Why? Is it a compat issue of 32-bit userland on 64-bit kernel?
32-bit userland on 64-bit kernel. kernel: sizeof(ulong) = 8 userspace: sizeof(ulong) = 4now, with big endian, a "1" is on the rightmost byte - which means looking at the bytes it's
kernel: byte[7] userspace: byte[3]So if you set bit nr "1" with the current logic, the kernel would set bit "1" (in the first 8 bytes), userspace would read bit "1" in the second byte, thus 32 + 1.
On little endian, the lower word is on the first 4 bytes, so it would still be bit "1" in the first byte.
Alex
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