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At 03:03 PM 5/10/2004, Andrew Urm wrote:<br>
<blockquote type=cite class=cite cite>Additional server instances don't
take up more RAM? Really? Or were you saying something
else?<br><br>
Andrew<br><br>
luke@techfreak.org wrote:<br><br>
<blockquote type=cite class=cite cite>Of course, there is a more
'elegant' way, but the guy is trying to<br>
get his feet wet here. So, I suggested an easy method.<br>
Also, I have run test after test and it doesnt take up any more RAM<br>
to run multiple installations, as opposed to using base_path. Or
are<br>
you taking about using disk space?<br>
Luke<br>
</blockquote></blockquote><br><br>
Additional instances do not take up more ram because the first
instance<br>
pulls the executable page code in to swap space in Unix/Linux. If
you<br>
have enough RAM, (and that is critical, or it Unix/Linux will need to
page<br>
swap, and there goes the lag game.. This falls on basic design theory
for<br>
Unix... You can never have enough ram. You always want to have enough
RAM<br>
to run your programs in ram, or you will page swap. Page swap =
slow,<br>
without getting in to the nasty details). How can I tell if I am
page swapping?<br><br>
run " vmstat". Check the man page for vmstat. You
will see a PI and PO,<br>
these should be zero. If not, you are swapping in pages of code to
execute.<br>
Add more ram, and you may have to relink the kernel, but most
Unix/Linux<br>
version support up to 4GB or more for 32 bit.<br><br>
The way it works is this.... Code for each game will execute in ram,
and<br>
share the page, but each will get their OWN data area's for Heap
and<br>
Stack.<br><br>
However, if one process (Game running) attempts to write to a code<br>
page, Linux will do a "Copy on Fork()". This is viewable in
vmstat<br>
as well. Basically it means that "A copy using the shared code
attempted<br>
to modify it, so it make a separate copy for the process that attempted
to<br>
write to the code base".<br><br>
Since this does not happen in COD, or most processes for that
matter,<br>
they all share the same single image in Swap to load to ram, and
the<br>
same single image of code in RAM. If you were to put the
processes<br>
in debug, and look at the registers, you would see the code segment<br>
registers are all the same, but the data pointer registers are all
different.<br><br>
Check out vmstat, and read the man page. It is worth your time.<br>
It will give you some good insight on how the resources are
allocated...<br><br>
Nice huh? Someday Windows will really work like that, or
maybe<br>
it will replace windows? {;-))<br><br>
MFrag<br>
{TRv}Team-Ravage<br>
aka. Mark J. DeFilippis, Ph. D EE<br><br>
<br>
<x-sigsep><p></x-sigsep>
<tt>-------------------------------------------------------------------------------<br>
Mark J.
DeFilippis
defilm@acm.org<br>
defilm@ieee.org<br><br>
<br>
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