Re: [PATCH] hw/riscv: boot: Reduce FDT address alignment constraints

[Bin Meng <bmeng.cn@gmail.com>]

+Atish

On Wed, Jun 8, 2022 at 2:20 PM Alistair Francis
<alistair.francis@opensource.wdc.com> wrote:
>
> From: Alistair Francis <alistair.francis@wdc.com>
>
> We previously stored the device tree at a 16MB alignment from the end of
> memory (or 3GB). This means we need at least 16MB of memory to be able
> to do this. We don't actually need the FDT to be 16MB aligned, so let's
> drop it down to 2MB so that we can support systems with less memory,
> while also allowing FDT size expansion.
>
> Resolves: https://gitlab.com/qemu-project/qemu/-/issues/992
> Signed-off-by: Alistair Francis <alistair.francis@wdc.com>
> ---
>  hw/riscv/boot.c | 4 ++--
>  1 file changed, 2 insertions(+), 2 deletions(-)
>
> diff --git a/hw/riscv/boot.c b/hw/riscv/boot.c
> index 57a41df8e9..e476d8f491 100644
> --- a/hw/riscv/boot.c
> +++ b/hw/riscv/boot.c
> @@ -226,11 +226,11 @@ uint64_t riscv_load_fdt(hwaddr dram_base, uint64_t mem_size, void *fdt)
>      /*
>       * We should put fdt as far as possible to avoid kernel/initrd overwriting
>       * its content. But it should be addressable by 32 bit system as well.
> -     * Thus, put it at an 16MB aligned address that less than fdt size from the
> +     * Thus, put it at an 2MB aligned address that less than fdt size from the
>       * end of dram or 3GB whichever is lesser.
>       */
>      temp = (dram_base < 3072 * MiB) ? MIN(dram_end, 3072 * MiB) : dram_end;
> -    fdt_addr = QEMU_ALIGN_DOWN(temp - fdtsize, 16 * MiB);
> +    fdt_addr = QEMU_ALIGN_DOWN(temp - fdtsize, 2 * MiB);
>

@Atish Patra  may have some pointers about the 16MiB alignment requirement.

Regards,
Bin