66

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!

Wuffers
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Ivan Xiao
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7 Answers7

38

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

To use it as sed instead of gsed, brew helpfully prints the following instructions after you install:

GNU "sed" has been installed as "gsed".
If you need to use it as "sed", you can add a "gnubin" directory
to your PATH from your bashrc like:

    PATH="/usr/local/opt/gnu-sed/libexec/gnubin:$PATH"

That is, append the following line to your ~/.bashrc or ~/.zshrc:

export PATH="/usr/local/opt/gnu-sed/libexec/gnubin:$PATH"
Ahmed Fasih
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33

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.

Lri
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13

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.]

Spiff
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8

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).

geekosaur
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7

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

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

results in

s/$/\
/g

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.

wisbucky
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3

This workaround works on Mac and the script can be executed also on Linux

NL="\n"
if [[ $uname -eq "Darwin" ]]; then
    NL=$'\\\n'
fi

echo 'foo bar' | sed -e "s| |${NL}|"
Massimo Zerbini
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0

A bit more compact form of @max-zerbini's answer:

[[ $(uname) -eq "Darwin" ]] && NL=$'\\\n' || NL="\n"
echo 'foo bar' | sed -e "s| |${NL}|"
slm
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