hw/riscv: boot: Reduce FDT address alignment constraints

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>
Reviewed-by: Atish Patra <atishp@rivosinc.com>
Reviewed-by: Bin Meng <bin.meng@windriver.com>
Tested-by: Bin Meng <bin.meng@windriver.com>
Message-Id: <20220608062015.317894-1-alistair.francis@opensource.wdc.com>
Signed-off-by: Alistair Francis <alistair.francis@wdc.com>

1 files changed, 2 insertions, 2 deletions

diff --git a/hw/riscv/boot.c b/hw/riscv/boot.c
index 2d80f40b31..06b4fc5ac3 100644
--- a/hw/riscv/boot.c
+++ b/hw/riscv/boot.c
@@ -227,11 +227,11 @@ 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);
 
     ret = fdt_pack(fdt);
     /* Should only fail if we've built a corrupted tree */