Sed: replacing newlines with "-z"?

Question

Problem: replace some regex with \n with sed.

Solution: there are many similar answers [1][2][3][4], and many other links that I won't link. All of them suggest you to create a new label :a, merge lines N, branch to :a if not end-of-file $!ba, and then do some command.

That said... In the GNU sed manual, there is the -z option:

-z
--null-data
--zero-terminated

Treat the input as a set of lines, each terminated by a zero byte
(the ASCII ‘NUL’ character) instead of a newline. This option can
be used with commands like ‘sort -z’ and ‘find -print0’ to process
arbitrary file names. 

So, first, for comparison reasons, if we try the naive approach:

$ seq 3 | sed 's/\n/ /g'
1
2
3

However, using this -z option:

$ seq 3 | sed -z 's/\n/ /g'
1 2 3

The Real Question: Why?

Given that it "merges" all the lines, as specified in the documentation, I expected that I would have to use \0 instead of \n, since:

Treat the input as a set of lines, each terminated by a zero byte (the ASCII ‘NUL’ character)

Since I didn't find any post related to it, I think I might be misunderstanding something here... So, what does it really do? Why does it work?

Answer 1

Using -z changes what sed considers to be a line. \n remains \n, but it doesn't end a line, but the null character (which is represented as \x0 in Sed) would. As there are no null bytes in the output of seq, the whole output is considered one line and processed in single iteration (i.e. replacing all \n's by spaces).