replace CR LF in text file using `sed` or `fart` (Find And Replace Text) - windows

Question

I have a 1.5 GB Windows text file with some lines ending with LF and most of lines ending with CR+LF
Can you please help with sed script which
will replace all CR+LF with $|$
replace all LF with CR+LF
replace back all $|$ with CR+LF
I have tried to do all replacements with text editor, but it took very long to perform all replacements in the file (1 percent for half an hour). I've tried to replace it with fart:
fart -c -B -b text.txt "\r\n" "$|$"
with following result
replacement 0 occurence(s) in 0 file(s)..

Answer 1

One with awk:
$ awk '{sub(/(^|[^\r])$/,"&\r")}1' file
Testing it (0x0a is LF, 0x0d is CR):
$ awk 'BEGIN{print "no\nyes\r\n\n\r"}' > foo
$ hexdump -C foo
00000000 6e 6f 0a 79 65 73 0d 0a 0a 0d 0a |no.yes.....|
0000000b
$ awk '{sub(/(^|[^\r])$/,"&\r")}1' foo > bar
$ hexdump -C bar
00000000 6e 6f 0d 0a 79 65 73 0d 0a 0d 0a 0d 0a |no..yes......|
0000000d

Answer 2

I would do this: first remove all \r at the end of the line, then explicitly add a \r to the end of the line.
sed -e 's/\r$//' -e 's/$/\r/' file
Here's a demo:
$ printf "1\r\n2\n3\n4\r\n5\n" > file
$ od -c file
0000000 1 \r \n 2 \n 3 \n 4 \r \n 5 \n
0000014
$ sed -i -e 's/\r$//' -e 's/$/\r/' file
$ od -c file
0000000 1 \r \n 2 \r \n 3 \r \n 4 \r \n 5 \r \n
0000017
This is GNU sed.

Answer 3

It's simpler just to install a util like unix2dos which does it automatically. With unix2dos the proposed intermediate step of converting CR+LF to $|$, (and back), isn't necessary. Demo:
# first dump a file with both *DOS* and *Unix* style line endings:
hexdump -C <({ seq 2 | unix2dos ; seq 3 4; } )
# the same file, run through unix2dos
hexdump -C <({ seq 2 | unix2dos ; seq 3 4; } | unix2dos)
Output:
00000000 31 0d 0a 32 0d 0a 33 0a 34 0a |1..2..3.4.|
0000000a
00000000 31 0d 0a 32 0d 0a 33 0d 0a 34 0d 0a |1..2..3..4..|
0000000c
Or more elaborately, a before/after table, (see man hexdump for details on formatting):
hdf() { hexdump -v -e '/1 "%_ad# "' -e '/1 " _%_u\_\n"' $# ; }
# Note: the `printf` stuff keeps `paste` from misaligning the output.
paste <(hdf <({ seq 2 | unix2dos ; seq 3 4; }) ; printf '\t\n\t\n' ; ) \
<(hdf <({ seq 2 | unix2dos ; seq 3 4; } | unix2dos ))
Output:
0# _1_ 0# _1_
1# _cr_ 1# _cr_
2# _lf_ 2# _lf_
3# _2_ 3# _2_
4# _cr_ 4# _cr_
5# _lf_ 5# _lf_
6# _3_ 6# _3_
7# _lf_ 7# _cr_
8# _4_ 8# _lf_
9# _lf_ 9# _4_
10# _cr_
11# _lf_