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Thanks for the reply
I just tried 'mem=96M' but free still reports 62748.
The following is taken from the /var/log/messages file :
Nov 19 08:03:20 kamco kernel: BIOS-provided physical RAM map:
Nov 19 08:03:20 kamco kernel: user-defined physical RAM map:
Nov 19 08:03:20 kamco kernel: 64MB LOWMEM available.
Nov 19 08:03:20 kamco kernel: Initializing CPU#0
Nov 19 08:03:20 kamco kernel: Memory: 62636k/66496k available (1733k kernel code
, 3472k reserved, 568k data, 112k init, 0k highmem)
Does anyone know what 'LOWMEM' means ??
and the syslog contains the following :
Nov 19 08:07:05 kamco kernel: Linux version 2.4.20 (root@midas) (gcc version 3.2
.2) #2 Mon Mar 17 22:02:15 PST 2003
Nov 19 08:07:05 kamco kernel: BIOS-88: 0000000000000000 - 000000000009f000 (usa
ble)
Nov 19 08:07:05 kamco kernel: BIOS-88: 0000000000100000 - 00000000040f0000 (usa
ble)
Nov 19 08:07:05 kamco kernel: user: 0000000000000000 - 000000000009f000 (usable
)
Nov 19 08:07:05 kamco kernel: user: 0000000000100000 - 00000000040f0000 (usable
)
Nov 19 08:07:05 kamco kernel: On node 0 totalpages: 16624
Nov 19 08:07:05 kamco kernel: zone(0): 4096 pages.
Nov 19 08:07:05 kamco kernel: zone(1): 12528 pages.
Nov 19 08:07:05 kamco kernel: zone(2): 0 pages.
Nov 19 08:07:05 kamco kernel: Kernel command line: auto BOOT_IMAGE=Linux ro root
=302 mem=96M ether=0,0,eth1
Does anyone understand this ? Why is BIOS-88 and user the same ?? (9F000 is 64 Meg I think ...). What does 40F000 refer to ? is this the HD ?
Last edited by simonburton; 11-19-2003 at 03:48 AM.
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