On Thu, Jan 16, 2014 at 08:39:06AM +1100, Peter Chubb wrote:
>
> In the multiboot header, there is a field, `mem_lower' that is meant to
> contain the size of memory starting at zero and ending below 640k.
> If your kernel is compiled with CONFIG_X86_RESERVE_LOW non zero
> (the usual case), then a hole is inserted into kernel's physical
> memory map at zero, so the test to find the size of this region in
> kexec/arch/i386/kexec-multiboot-x86.c never succeeds, so the value is
> always zero.
>
> On a PC99 architecture, there is always memory at physycal address zero;
> assume that a region that starts below 64k actually starts at zero,
> and use it for the mem_lower variable.
>
> Signed-off-by: Peter Chubb <peter.chubb at nicta.com.au>
Thanks Peter, I have applied this.
>
> ---
> kexec/arch/i386/kexec-multiboot-x86.c | 14 +++++++++++---
> 1 file changed, 11 insertions(+), 3 deletions(-)
>
> Index: kexec-tools-2.0.4/kexec/arch/i386/kexec-multiboot-x86.c
> ===================================================================
> --- kexec-tools-2.0.4.orig/kexec/arch/i386/kexec-multiboot-x86.c 2013-03-14 18:45:16.000000000 +1000
> +++ kexec-tools-2.0.4/kexec/arch/i386/kexec-multiboot-x86.c 2014-01-15 10:21:02.138172304 +1000
> @@ -261,10 +261,18 @@ int multiboot_x86_load(int argc, char **
> mmap[i].length_high = length >> 32;
> if (range[i].type == RANGE_RAM) {
> mmap[i].Type = 1; /* RAM */
> - /* Is this the "low" memory? */
> - if ((range[i].start == 0)
> - && (range[i].end > mem_lower))
> + /*
> + * Is this the "low" memory? Can't just test
> + * against zero, because Linux protects (and
> + * hides) the first few pages of physical
> + * memory.
> + */
> +
> + if ((range[i].start <= 64*1024)
> + && (range[i].end > mem_lower)) {
> + range[i].start = 0;
> mem_lower = range[i].end;
> + }
> /* Is this the "high" memory? */
> if ((range[i].start <= 0x100000)
> && (range[i].end > mem_upper + 0x100000))
>
> --
> Dr Peter Chubb peter.chubb AT nicta.com.au
> http://www.ssrg.nicta.com.au Software Systems Research Group/NICTA
>
