[ncolug] Re: some bash to shorten
- From: nor thern <zboson2003@xxxxxxxxx>
- To: ncolug@xxxxxxxxxxxxx
- Date: Sun, 29 Nov 2009 22:46:37 -0800 (PST)
thank you , Chuck.
the outputs are shown below.
ls -l /proc/kcore | cut -f5 -d" "
9663680512
grep "MemTotal" /proc/meminfo|cut -d ":" -f2|sed '/ kB/ s/ kB//g'|cut -c9-
8199584
free | grep "Mem"
Mem: 8199584 5968036 2231548 <snipped off last fields>
tried to factor the proc/kcore value to see where it came from.
2^12 x 61 x 38677
does /proc/kcore include video ram to make the number larger?
--- On Sun, 11/29/09, cstickelman@xxxxxxxxxx <cstickelman@xxxxxxxxxx> wrote:
From: cstickelman@xxxxxxxxxx <cstickelman@xxxxxxxxxx>
Subject: [ncolug] Re: some bash to shorten
To: ncolug@xxxxxxxxxxxxx
Cc: "nor thern" <zboson2003@xxxxxxxxx>
Date: Sunday, November 29, 2009, 9:04 PM
How about:
ls -l /proc/kcore | cut -f5 -d" "
or
ls -lh /proc/kcore | cut -f5 -d" "
if you want the results to be human-readable...
Chuck
---- nor thern <zboson2003@xxxxxxxxx> wrote:
> grep "MemTotal" /proc/meminfo|cut -d ":" -f2|sed '/ kB/ s/ kB//g'|cut -c9-
>
> this is the shortest that i have found to print the RAM size as an integer.
> unprivileged user .
> suggestions to shorten this appreciated.
>
>
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