26%
11.02.2016
+----------------------------------+---------+---------------------+
40 | 47e0142a3638fdc24fe40d4e4fbce3f1 | Row 1 | 2015-09-13 15:24:12 |
41 | b833c1e4c5bfc47d0dbe31c2e3f30837 | Row 3 | 2015-09-13 15:24:14 |
42 | c7d46523a316de4e1496c65c3cbdf358 | Row 2 | 2015
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30.11.2025
, appearing in alphabetical order but allowing intervening letters, you can use the search expression:
"a.*e.*i.*o.*u"
This would match lines 1, 2, and 3. If you want lines containing all five vowels in order
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30.11.2025
and modifying the IP address in line 8. The line
lxc-create -n guest -f /lxc/conf.guest
Listing 3
Container Configuration: conf.guest
01 lxc.utsname = guest
02 lxc.tty = 4
03 lxc
26%
05.02.2023
wgossrvc
Listing 4
setup_goss.sh Diff Additions
5a6
> GOSSVFLE='/lib/systemd/system/goss.service'
9,10c10,12
< sha256sum
< tee
---
> sha256sum
> systemctl
> tee
81a84
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30.11.2025
creates a 256MB file in the current directory along with process for the job. This process reads complete file content in random order. Fio records the areas that have already been read and reads each area
26%
30.11.2025
filesystem, if so desired [5]:
$ guestfish -N fs:ext3
When you call Guestfish with the new image,
$ guestfish -a
you are taken to a shell where you first need to enter run to toggle the system ... 9
26%
30.11.2025
); i+= 4096) newblock[i] = 'Y';
12 printf("Allocated %d MB\n", allocation);
13 }
14 }
Things are more interesting when memory is being used. Uncommenting line 11 does just that. The OOM
25%
07.06.2019
:
> numbers <- c(1, 2, 3, 4, 5)
The c() function – the c stands for "concatenate" – combines the individual elements listed in parentheses. An equals sign can be used as an alternative for assignments, in line
25%
10.06.2015
to the first sed I know that here is only a SINGLE space
45 display_list="$(sed ':a;N;$!ba;s/\n / /g'<<<"$xrandr_current" | sed \
-n -e 's/^\([a-zA-Z0-9_-]\+\) connected.* \([0-9]\+\)mm.* \([0-9]\+\)mm
25%
28.07.2025
[i] + b[i];
}
When the number of cycles is known at compile time, a loop can be fully unrolled:
c[0] = a[0] + b[0];
c[1] = a[1] + b[1];
c[2] = a[2] + b[2];
c[3] = a[3] + b[3];
However, it remains
