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I find that \n doesn't work in sed under Mac OS X. Specifically, say I want to break the words separated by a single space into lines:

# input
foo bar

I use,

echo "foo bar" | sed 's/ /\n/'

But the result is stupid, the \n is not escaped!

foonbar

After I consulted to google, I found a workaround:

echo 'foo bar' | sed -e 's/ /\'$'\n/g'

After reading the article, I still cannot understand what \'$'\n/g' means. Can some one explain it to me, or if there is any other way to do it? Thanks!

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this would probably work too: echo "foo bar" | tr ' ' '\n' –  glenn jackman Jul 6 '11 at 18:17
1  
Thanks for the advice. But currently I just use the above case as an example, I do need to know how to escape a \n. –  Ivan Z. G. Xiao Jul 6 '11 at 22:50
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4 Answers

These would also work:

echo 'foo bar' | sed 's/ /\
/g'

echo 'foo bar' | sed $'s/ /\\\n/g'

lf=$'\n'; echo 'foo bar' | sed "s/ /\\$lf/g"

OS X's sed doesn't interpret \n in the replace pattern, but you can use a literal linefeed preceded by a line continuation character. The shell replaces $'\n' with a literal linefeed before the sed command is run.

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You can brew install gnu-sed and replace calls to sed to gsed.

If you don't want to prepend the "g" to sed, you can brew install gnu-sed --default-names and just call sed.

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The expression $'...' is a bash-ism which produces ... with the standard escape sequences expanded. Th \' before it just means a backslash followed by the end of the quoted section, the resulting string is s/ /\. (Yes, you can switch quoting in the middle of a string; it doesn't end the string.)

POSIX standard sed only accepts \n as part of a search pattern. OS X uses the FreeBSD sed, which is strictly POSIX compliant; GNU, as usual, adds extra stuff and then Linux users all think that is some kind of "standard" (maybe I'd be more impressed if either of them had a standards process).

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Now I understand the $'...' part. But... what is s/ /\ ? What do you mean by switch quoting? –  Ivan Z. G. Xiao Jul 6 '11 at 22:53
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'...' is one kind of shell quoting; $'...' is another. There's also "..." and \x to quote a single character. You can combine those in a single word, which is what was being done there, switching from a normal '' string to a $'' string to translate the \n. As for the rest, it's building up a sed command (s/text/replacement/flags). In this case the command is started, including a backslash at the end to protect the literal newline that the $'\n/g' appends. The result is to replace all (the /g flag) spaces with newlines. –  geekosaur Jul 6 '11 at 23:01
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The workaround you found passes a single argument string to sed -e.

That argument ends up being a string in the familiar sed s/ / /g format.

That string is created in two parts, one after the other.

The first part is quoted in '...' form.

The second part is quoted in $'...' form.

The 's/ /\' part gets the single-quotes stripped off, but otherwise passes through to sed just as it looks on the command-line. That is, the backslash isn't eaten by bash, it's passed to sed.

The $'\n/g' part gets the dollar sign and the single-quotes stripped off, and the \n gets converted to a newline character.

All together, the argument becomes

s/ /\newline/g

[That was fun. Took a while to unwrap that. +1 for an interesting question.]

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