Re: [MPlayer-users] Optimization for a slow CPU

[Grant <emailgrant@xxxxxxxxx>]

> "I don't know about video memory." The instruction on
> the mplayer link that I mentioned is very clear about
> how to know the video ram base addr and its size. You
> ought to get those numbers and then run that "ech0"
> command with them as values.
>
> My cat /proc/mtrr looks like this:
>
> reg00: base=0x00000000 (   0MB), size= 512MB: write-back, count=1
> reg01: base=0xf0000000 (3840MB), size= 128MB: write-combining, count=1
>
>
> I don't have any memory sticks, only 512MB of main mem and
> max 128MB of shared video ram.

Sorry about that.  I'm seeing three different values related to video RAM.

Here's my cat /proc/mtrr :

reg00: base=0x00000000 (   0MB), size= 128MB: write-back, count=1
reg01: base=0x08000000 ( 128MB), size=  64MB: write-back, count=1
reg02: base=0x0bf00000 ( 191MB), size=   1MB: uncachable, count=1

the output of dmesg :

agpgart: Detected an Intel i810 E Chipset.
agpgart: detected 4MB dedicated video ram.
agpgart: AGP aperture is 64M @ 0xf8000000

and the output of lspci -v :

Memory at f8000000 (32-bit, prefetchable) [size=64M]
Memory at ffa80000 (32-bit, non-prefetchable) [size=512K]

I know I have 192MB of system RAM (128+64).  dmesg says 4MB and 64MB
in the context of video RAM.  lspci -v says 64MB and 512KB.  Should I
execute the following based on 4MB:

echo "base=0xf8000000 size=0x400000 type=write-combining" >| /proc/mtrr

- Grant
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