On 11 November 2011 15:59, Alexander Graf<address@hidden> wrote:
This is the machine init function. The s390 virtio machine's ram layout is
defined to be exactly as I posted in the previous post. So there won't be
any ram starts at != 0 or multiple mappings :).
That's the layout of the RAM in target_phys_addr_t space.
The layout of the ram in ram_addr_t space is not guaranteed
to be the same, it just happens to be in this case.
On 10.11.2011, at 02:36, Peter Maydell wrote:
In summary, either:
(1) you need to ask the memory region for its ram_addr_t
[there doesn't seem to be an API for this at the moment],
do arithmetic on ram_addr_t's and use qemu_get_ram_ptr() to
get a host pointer corresponding to that ram_addr_t
or (2) you need to do your arithmetic in target_phys_addr_t's
(and then you can say your RAM starts at physaddr 0, which
is guaranteed, rather than at ram_addr_t 0, which isn't)
and use cpu_physical_memory_map/unmap().
I still don't get it. What I want is:
for (i = ram_size; i< my_ram_size; i++)
stb_phys(env, i, 0);
cpu_physical_memory_map gives me a pointer to the memory region between
ram_size and my_ram_size. I memset(0) it. I don't see how I could possibly
have different offsets anywhere. If cpu_physical_memory_map would take
anything different than physical address, a lot of assumptions in QEMU code
would break.
Yes, so that's option (2) and you need to be using a target_phys_addr_t.