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I've a long hex string 00121eafc5800020 ... etc.

Is there an easy way I can split these with a space per hex coded byte so I get 00 12 1e af c5 80 00 20 in vim (or another easily accessible tool)

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up vote 5 down vote accepted
$ sed 's/\(..\)/\1 /g' <<< '1234567890'
12 34 56 78 90 
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Same command works in vim too, just add a line number, a range, or a % in front of it as appropriate. (Uh, I'll mention that you should remove the 'sed' bit... just in case, I've seen some impressive people here.) – lornix Oct 21 '11 at 11:18
Which is equivalent to s/../& /g – grawity Oct 21 '11 at 12:42
note that the /g at the end will dice up everything on the given line. – Conrad.Dean Nov 2 '11 at 2:46
How does that regular expression work? – Tom Wijsman Nov 19 '11 at 19:51
It puts 2 characters in group 1, then outputs group 1 with a space after it. – Ignacio Vazquez-Abrams Nov 19 '11 at 20:23

Try this key sequence (from the start of the number): q1lxPpr lq@1@1@1@1@1@1@1@1 (etc. or say, 20@1)

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"Try this key sequence" without any explanation is never a good answer to vim questions. – grawity Oct 21 '11 at 12:43
Maybe. But that would spoil the fun of finding out. – Benedict Oct 21 '11 at 12:54
Last I checked, this site was where people went for good answers. – grawity Oct 21 '11 at 12:56
I agree. This is a good answer. You'll note the (also good) accepted answer is also pretty terse. If you want to edit both to add a wordy explanation you are welcome to. Alternatively you can trust that the average reader knows enough to lookup the details to sed regex or the vim key map. – Benedict Oct 21 '11 at 13:06

In vim, you can use :s/\([0-9a-f][0-9a-f]\)/\1 /g to do what you want. To test this, you can use the following command:

dd bs=512 if=/dev/urandom count=2 2>/dev/null | xxd -p | tr -d '\n' | vim -

Then in vim, use :s/\([0-9a-f][0-9a-f]\)/\1 /g to add a space after every pair of hex-characters. Here's a small example:

$ dd bs=32 if=/dev/urandom count=1 2>/dev/null | xxd -p | tr -d '\n' | vim -
:s/\([0-9a-f][0-9a-f]\)/\1 /g
6a ad 5f 80 0a 73 9f ad bc 3e 23 12 23 0d ae ca 80 55 49 15 77 b0 e0 d7 5a 37 bb 37 77 f0 a3 e0 
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:%s/\x\x/& /g

is enough.

If you want to cut into several lines, you can also apply this :substitute first.


Or if you prefer a one-liner (I'm not sure this is a better solution here)

:%s/\x\{16}/\=substitute(submatch(0), '\x\x', '& ', 'g')."\n"/g
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