In normal windows-to-unix conversion, you can do something like sed s/\r//g, which removes the \r characters from the stream.
But I'm trying to convert endlines of files that could be mac encoded (\r) or windows encoded (\r\n). So I cannot just remove the \r, as it would delete the mac endings if there's any. I have to "canonicalize" the line-ending characters first. This canonicalization step converts from \r\n to \r (after which I do the \r to \n conversion). Yet, I'm not able to solve this step with sed. I tried something like this:
$> echo -e "foo\r\nbar" | sed 's/\r\n/\r/g' | xxd -c 24 -g 1
00000000: 66 6f 6f 0d 0a 62 61 72 0a foo..bar.
I was able to solve it with bbe like this:
$> echo -e "foo\r\nbar" | bbe -e 's/\r\n/\r/g' | xxd -c 24 -g 1
00000000: 66 6f 6f 0d 62 61 72 0a foo.bar.
Is it possible to do the same with sed?
sed by default splits input on \n, so \n never ends up in the pattern space. However, if you are using GNU sed, you can use -z/--null-data option to make sed treat the input as NUL character separated lines:
$ echo -e "foo\r\nbar" | sed -z 's/\r\n/\r/g' | hd
00000000 66 6f 6f 0d 62 61 72 0a |foo.bar.|
Alternatively, in POSIX sed, you can append all lines to the pattern space (with N command in a loop), effectively copying the complete file to the pattern space, and then do the substitute:
$ echo -e "foo\r\nbar" | sed -n ':a;N;ta; s/\r\n/\r/g; p' | hd
00000000 66 6f 6f 0d 62 61 72 0a |foo.bar.|