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    <p>I think <code>awk</code> is probably the better tool for this job, but it can be done with <code>sed</code>:</p> <pre><code>sed '/^[^ ]* *00 *|/{ :a s/\(|.*[^(]\)\([0-9a-f][1-9a-f]\)/\1(\2)/ t a :b s/\(|.*[^(]\)\([1-9a-f][0-9a-f]\)/\1(\2)/ t b }' data </code></pre> <p>The script looks for lines containing <code>00</code> before the pipe, and only applies the operations to those lines. There are two substitute operations, each wrapped in a loop. The <code>:a</code> and <code>:b</code> lines are labels. The <code>t a</code> and <code>t b</code> commands are a conditional jump to the named label if there was a substitution performed since the last jump. The two substitutions are almost symmetric; the first deals with any number not ending in 0; the second deals with any number not starting with 0; between them, they ignore <code>00</code>. The patterns look for a pipe, any sequence of characters not ending with an open parenthesis <code>(</code>, and the appropriate pair of digits; it replaces that so that the number ends up inside parentheses. The loops are necessary because a <code>g</code> modifier doesn't start from the beginning again, and the patterns work backwards through the numbers.</p> <p>Given this data file (a slightly extended version of yours):</p> <pre><code>foobar 42 | ff 00 00 00 00 foobaz 00 | 0a 00 0b 00 00 foobie 00 | 00 00 00 00 00 bar 00 | ab ba 00 cd 00 fizbie 00 | ab ba 00 cd 90 fizzbuzz 00 | ab ba 00 cd 09 </code></pre> <p>the output from the script is:</p> <pre><code>foobar 42 | ff 00 00 00 00 foobaz 00 | (0a) 00 (0b) 00 00 foobie 00 | 00 00 00 00 00 bar 00 | (ab) (ba) 00 (cd) 00 fizbie 00 | (ab) (ba) 00 (cd) (90) fizzbuzz 00 | (ab) (ba) 00 (cd) (09) </code></pre> <p>It is moderately educational to add a <code>p</code> after each of the substitute commands, so you can see how the substitutions work:</p> <pre><code>foobar 42 | ff 00 00 00 00 foobaz 00 | 0a 00 (0b) 00 00 foobaz 00 | (0a) 00 (0b) 00 00 foobaz 00 | (0a) 00 (0b) 00 00 foobie 00 | 00 00 00 00 00 bar 00 | ab ba 00 (cd) 00 bar 00 | ab (ba) 00 (cd) 00 bar 00 | (ab) (ba) 00 (cd) 00 bar 00 | (ab) (ba) 00 (cd) 00 fizbie 00 | ab ba 00 (cd) 90 fizbie 00 | ab (ba) 00 (cd) 90 fizbie 00 | (ab) (ba) 00 (cd) 90 fizbie 00 | (ab) (ba) 00 (cd) (90) fizbie 00 | (ab) (ba) 00 (cd) (90) fizzbuzz 00 | ab ba 00 cd (09) fizzbuzz 00 | ab ba 00 (cd) (09) fizzbuzz 00 | ab (ba) 00 (cd) (09) fizzbuzz 00 | (ab) (ba) 00 (cd) (09) fizzbuzz 00 | (ab) (ba) 00 (cd) (09) </code></pre>
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    1. POSed: Complicated replace after pattern (on same line)
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    1. USJonathan Leffler
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    1. COAha, that looping construct was the missing piece of the puzzle! Good information to remember. I can rest easy knowing that sed can indeed accomplish the task, albeit a bit round-a-bout.
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    2. COYes, the `awk` solution is neater, though. I suspect there are other ways to do the `awk` solution, but they're all neater than `sed`. The looping construct is powerful, though. In fact, I read somewhere (from a link on SO) that [`sed` is Turing complete](http://www.catonmat.net/blog/proof-that-sed-is-turing-complete/) — but it is not a good choice for a general purpose language.
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      1. USJonathan Leffler
    3. COthough two leading spaces from the input are removed ;)
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