Re: [PATCH 07/12] i386: Add missing !X86_PAE dependincy to the 2G/2G split.



"H. Peter Anvin" <hpa@xxxxxxxxx> writes:

Eric W. Biederman wrote:
When in PAE mode we require that the user kernel divide to be
on a 1G boundary. The 2G/2G split does not have that property
so require !X86_PAE

?????

-hpa

From arch/i386/Kconfig:

choice
depends on EXPERIMENTAL
prompt "Memory split" if EMBEDDED
default VMSPLIT_3G
help
Select the desired split between kernel and user memory.

If the address range available to the kernel is less than the
physical memory installed, the remaining memory will be available
as "high memory". Accessing high memory is a little more costly
than low memory, as it needs to be mapped into the kernel first.
Note that increasing the kernel address space limits the range
available to user programs, making the address space there
tighter. Selecting anything other than the default 3G/1G split
will also likely make your kernel incompatible with binary-only
kernel modules.

If you are not absolutely sure what you are doing, leave this
option alone!

config VMSPLIT_3G
bool "3G/1G user/kernel split"
config VMSPLIT_3G_OPT
depends on !HIGHMEM
bool "3G/1G user/kernel split (for full 1G low memory)"
config VMSPLIT_2G
depends on !X86_PAE
bool "2G/2G user/kernel split"
config VMSPLIT_1G
bool "1G/3G user/kernel split"
endchoice

config PAGE_OFFSET
hex
default 0xB0000000 if VMSPLIT_3G_OPT
default 0x78000000 if VMSPLIT_2G
default 0x40000000 if VMSPLIT_1G
default 0xC0000000

The default PAGE_OFFSET is at 0x7800000 for the 2G/2G split.
All of these options were originally !X86_PAE, I assume
the intention was to be able to have 2G of RAM without
needing high memory.

I don't really care I just saw the problem and decided to prevent
people trying a combination that simply doesn't work.

Eric
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