How do I use a new-line replacement in a BSD sed?

Question

Greetings, how do I perform the following in BSD sed?

sed 's/ /\n/g'

From the man-page it states that \n will be treated literally within a replacement string, how do I avoid this behavior? Is there an alternate?

I'm using Mac OS Snow Leopard, I may install fink to get GNU sed.

Stoical

Small correction: \n isn't treated literally as such, it gets translated to literal n (i.e., the `` is dropped; but that obviously still isn't the desired outcome). – Consummation

Answer 1

In a shell, you can do:

    sed 's/ /\
/g'

hitting the enter key after the backslash to insert a newline.

Elysian

Fantastic, thanks Wooble. Is there any known reason why BSD sed doesn't support \n in the replacement string? Just curious. – Stoical

I'm not sure, but it seems they just decided to not support escapes at all in the replacement string. sed 'y/ /\n/' seems to work, although it only does string replacement; if you need to replace the ' ' with a regex it will fail. Also, the \n seems to be special-cased; \r with get you an 'r', not a carriage return. – Elysian

Answer 2

Another way:

sed -e 's/ /\'$'\n/g'

See here.

Carbolated

+1, but note that this relies on a shell feature, ANSI C quoting ($'...'), and some shells do not support it (e.g., dash - it's not part of POSIX). Also,'s/ /\'$'\n''/g' - wedging a $'\n' between two single-quoted strings - is conceptually clearer and more robust. Alternatively, use $'s/ /\\\n/g' (note the need to escape the \ ). – Consummation

Answer 3

For ease of use, i personally often use

cr="\n" 
# or (depending version and OS)
cr="
"

sed "s/ /\\${cr}/g"

so it stays on 1 line.

Optional

Interesting; can I suggest calling the variable nl rather than cr? Assuming you use a shell that supports ANSI C quoting (e.g., bash, ksh, zsh), you could simplify the assignment to cr=$'\n'; then you wouldn't have to worry about platform differences. – Consummation

cool advice, sur about the $'\n' on any "unix" ? it will indeed help (my Sun give me some issue on assignation/action like that with \n) – Optional

$'\n' (ANSI-C quoting) should work if you have the right shell (bash, ksh, or zsh - NOT in POSIX-features-only shells such as dash, however), so the platform should not matter. (bash introduced the feature in version 2.0 (a loooong time ago) - not sure about ksh, and zsh, but I would expect current systems to have recent-enough versions). – Consummation

In NetBSD 9.3, this gives me the error unterminated substitute in regular expression :-( I genuinely cannot find a way to make sed use newlines in NetBSD, but you could use awk -v RS=" " "{print}" instead, taken from this answer – Woothen

Answer 4

To expand on @sikmir's answer: In Bash, which is the default shell on Mac OS X, all you need to do is place a $ character in front of the quoted string containing the escape sequence that you want to get interpreted. Bash will automatically translate it for you.

For example, I removed all MS-DOS carriage returns from all the source files in lib/ and include/ by writing:

grep -lr $'\r' lib include | xargs sed -i -e $'s/\r//'
find . -name '*-e' -delete

BSD grep would have interpreted '\r' correctly on its own, but using $'\r' doesn't hurt.

BSD sed would have misinterpreted 's/\r//' on its own, but by using $'s/\r//', I avoided that trap.

Notice that we can put $ in front of the entire string, and it will take care of all the escape sequences in the whole string.

$ echo $'hello\b\\world'
hell\world

Ungulate