Using sed to split hex string

UNIX for Beginners Q & A
1

Hi,

I'm looking to split the following hex string into rows of four elements.

I've tried the following but it doesn't seem to work. How can I tell sed to match based on a pair of number(s) and letter(s), and add a newline every 4 pairs?
In addition, I need to add another newline after every string has been processed.

echo "1C 50 A9 D3 8B B0 24 9C 14 1C 50 A9 D3 8B B0 24" | sed -nr 's/[0-9A-Z]{2}/\n/4p'
1C 50 A9
 8B B0 24 9C 14 1C 50 A9 D3 8B B0 24 9C 14

Expected output:

1C 50 A9 D3
8B B0 24 9C

14 1C 50 A9
D3 8B B0 24
2 
 | sed 's/\(\([0-9A-F][0-9A-F] \)\{4\}\)/\1\n/g'
3

Insert a \n every 12 characters:

sed -nr 's/.{12}/&\n/gp'

If you really want to indentify 4 elements:

sed -nr 's/([0-9A-F]{2} ){4}/&\n/gp'

With BRE (no -r) it is like the previous solution:

sed -n 's/\([0-9A-F][0-9A-F] \)\{4\}/&\n/gp'
1 Like
4

regex / wctype offer a "character class" for hex:

sed -nr 's/([[:xdigit:]]{2} ){4}/&\n/gp'
1 Like
5

Hi.

echo "1C 50 A9 D3 8B B0 24 9C 14 1C 50 A9 D3 8B B0 24" |
fold -12 |
sed '2~2a\
'

producing:

1C 50 A9 D3 
8B B0 24 9C 

14 1C 50 A9 
D3 8B B0 24

Where the GNU extension '2~2' starts with line 2, then after each 2nd line adds a newline.

On a system like:

OS, ker|rel, machine: Linux, 3.16.0-7-amd64, x86_64
Distribution        : Debian 8.11 (jessie) 
bash GNU bash 4.3.30
fold (GNU coreutils) 8.23
sed (GNU sed) 4.2.2

Best wishes ... cheers, drl

6

You might try as well

echo "1C 50 A9 D3 8B B0 24 9C 14 1C 50 A9 D3 8B B0 24" | tr ' ' $'\n' | paste -sd'   \n'