Norbert Cartagena wrote:
>
> Actually what you're seeing is pretty normal. Here's a
> short explanation about what's very possibly going on
> (I don't know MUCH about this, but I'm sure someone
> can expand upon this). The way the kernel works when
Briefly:
Linux (and other modern UNIX-like systems) utilize a unified memory/file
cache VM system. Whenever you read anything from or write anything to a
block device (i.e. hard drive, floppy drive, cdrom drive) those blocks
are also cached in memory in case you need to use them again any time
soon.
Also, as gNorb pointed out, when programs get loaded and executed, the
various libraries and system "things" that get loaded into memory also
sit around even after they've stopped being used.
Like Norb outlined, when stuff like that is in memory, that amount of
memory shows up as "used." That memory stays allocated until memory
pressure forces the VM to start reclaiming blocks. Typically the first
thing to go are buffer cache blocks - i.e. stuff cached from block
devices. The thinking is that you can always re-read them off disk if
necessary. Only when you've reduced the amount of buffer cache to a
threshhold minimum value (settable in /proc/sys someplace) will the VM
start to swap program pages out to disk, thereby freeing up more
physical RAM.
What's kind of interesting about this approach is that the same memory
pages are shared between the buffer cache and program memory. In the
olden DOS and Windows 3.x days, stuff like "smartdrive" used to be
required to pre-allocate a certain amount of memory exclusively for
caching disk reads/writes. The idea of a unified memory space is fairly
new, and I'm not entirely convinced that even "modern" versions of
Windows utilize this kind of VM system since I've seen Windows boxes
with 4GB of RAM thrash hard drives under even moderate reads/writes as
long as you keep reading/writing enough different data.
So how much memory your kernel reports as "used" is entirely up to how
much stuff you've read/written to disk lately. It's entirely possible
that even with 2GB of RAM, you can actually show all 2GB of RAM being
"used" even when you have no programs in memory because all of that
memory has been allocated for buffer cache or program cache, even though
it's not in use anymore.
If you want to see how much memory programs are actually taking, use the
command:
ps axv
and that'll show you various memory-related statistics about what's
currently running.
--
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
#!/usr/bin/perl -w
$_='while(read+STDIN,$_,2048){$a=29;$b=73;$c=142;$t=255;@t=map
{$_%16or$t^=$c^=($m=(11,10,116,100,11,122,20,100)[$_/16%8])&110;
$t^=(72,@z=(64,72,$a^=12*($_%16-2?0:$m&17)),$b^=$_%64?12:0,@z)
[$_%8]}(16..271);if((@a=unx"C*",$_)[20]&48){$h=5;$_=unxb24,join
"",@b=map{xB8,unxb8,chr($_^$a[--$h+84])}@ARGV;s/...$/1$&/;$d=
unxV,xb25,$_;$e=256|(ord$b[4])<<9|ord$b[3];$d=$d>>8^($f=$t&($d
>>12^$d>>4^$d^$d/8))<<17,$e=$e>>8^($t&($g=($q=$e>>14&7^$e)^$q*
8^$q<<6))<<9,$_=$t[$_]^(($h>>=8)+=$f+(~$g&$t))for@a[128..$#a]}
print+x"C*",@a}';s/x/pack+/g;eval
usage: qrpff 153 2 8 105 225 < /mnt/dvd/VOB_FILENAME \
| extract_mpeg2 | mpeg2dec -
http://www.cs.cmu.edu/~dst/DeCSS/Gallery/
http://www.eff.org/ http://www.anti-dmca.org/
http://www.sciencemag.org/cgi/content/full/293/5537/2028
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