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!


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!

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

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

(Edit: updated brew flag, hat tip to Clément.)

  • Using the same software on all platform is way easier (at last to me) than dealing with every specificities of the Mac version. Beware, the option is now --with-default-names and not --default-names. However, this option did not worked on my installation, so I had to put a alias gsed=sed in my ~/.profile to make it work. – Clément May 11 '15 at 11:40
  • 4
    This is an ugly workaround. It doesn't explain why sed on OS X behaves the way it does and makes the incorrect assumption that gnu-sed is more correct. Don't be a GNU-addict and stick to POSIX standards to avoid problems in the long run. – octosquidopus Oct 8 '15 at 2:13
  • It is how Apple should make its OS works by default. Why use the name of a program and then make it works differently? I waste an hour because of this... – rascio Apr 4 '17 at 10:58
  • @rascio - because OS X is BSD-like and Linux is Linux-like. – iAdjunct May 30 at 14:30
  • @rascio I think you'll find that GNU sed is the newcomer; BSD sed dates back to 1979 (see en.wikipedia.org/wiki/Version_7_Unix). – Duncan Bayne Sep 10 at 5:19

These would also work:

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

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.

  • 4
    Why does sed $'s/ /\\\n/g' work, but not sed $'s/\\\n/ /g'? – Alec Jacobson Jan 3 '16 at 23:03
  • 1
    @alec You can't use sed even in linux/unix to remove newlines because its parsed/split at each newline. If you run this in linux/unix, it won't do anything either: echo -e 'foo\nbar' | sed 's/\n//' – wisbucky Oct 27 '17 at 1:05

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.]

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).

  • 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
  • 2
    '...' 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

There's a very easy to visually see what's happening. Simply echo the string!

echo 's/$/\'$'\n/g'

results in


which is equivalent to s/$/\newline/g

If you didn't have the extra \ before the newline, the shell would interpret the newline as the end of the command prematurely.

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