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>
---
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
