Re: interpret this BIOS

[reg dwf com]

> On Sat, May 15, 2004 at 11:51:16AM -0600, reg dwf com wrote:

> > BIOS-provided physical RAM map:
> >  BIOS-e820: 0000000000000000 - 000000000009e800 (usable)
> 
> Most of 640K RAM
> >  BIOS-e820: 000000000009e800 - 00000000000a0000 (reserved)
> 
> A bit reserved (EBDA)
> 
> >  BIOS-e820: 00000000000f0000 - 0000000000100000 (reserved)
> 
> ROMs
> 
> >  BIOS-e820: 0000000000100000 - 000000002fffc000 (usable)
> Lots of memory
> 
> >  BIOS-e820: 000000002fffc000 - 000000002ffff000 (ACPI data)
> >  BIOS-e820: 000000002ffff000 - 0000000030000000 (ACPI NVS)
> 
> Two chunks used by ACPI
> 
> >  BIOS-e820: 00000000ffff0000 - 0000000100000000 (reserved)
> 
> Ending a 4Gb
> 
> 
> -- 

Perhaps you can be a bit more specific.
The system in fact has a 512MB chip and a 256MB chip for a total of 768/775Mb

You seem to show that this is shown in BIOS as a single entry, rather
than an entry for each stick, is that true?

And of course, I dont understand how to interpret the numbers that I see.
My first guess would be that they are starting point, size (is there any
documentation?,- I didnt find it)

But if I believe that, then 0x2fffc000 = 805289984 which is wrong for the size.

So, back to my initial question: How do I interpret this stuff?

-- 
                                        Reg.Clemens
                                        reg dwf com