How to insert a newline in front of a pattern?

This works in bash and zsh, tested on Linux and OS X:

sed 's/regexp/\'$'\n/g'

In general, for $ followed by a string literal in single quotes bash performs C-style backslash substitution, e.g. $'\t' is translated to a literal tab. Plus, sed wants your newline literal to be escaped with a backslash, hence the \ before $. And finally, the dollar sign itself shouldn't be quoted so that it's interpreted by the shell, therefore we close the quote before the $ and then open it again.

Edit: As suggested in the comments by @mklement0, this works as well:

sed $'s/regexp/\\\n/g'

What happens here is: the entire sed command is now a C-style string, which means the backslash that sed requires to be placed before the new line literal should now be escaped with another backslash. Though more readable, in this case you won't be able to do shell string substitutions (without making it ugly again.)

Some of the other answers didn't work for my version of sed.Switching the position of & and \n did work.

sed 's/regexp/\n&/g' 

Edit: This doesn't seem to work on OS X, unless you install gnu-sed.

In sed, you can't add newlines in the output stream easily. You need to use a continuation line, which is awkward, but it works:

$ sed 's/regexp/\&/'

Example:

$ echo foo | sed 's/.*/\&/'foo

See here for details. If you want something slightly less awkward you could try using perl -pe with match groups instead of sed:

$ echo foo | perl -pe 's/(.*)/\n$1/'foo

$1 refers to the first matched group in the regular expression, where groups are in parentheses.