Why does "sed -n -i" delete existing file contents?

Question

Running Fedora 25 server edition. sed --version gives me sed (GNU sed) 4.2.2 along with the usual copyright and contact info. I've create a text file sudo vi ./potential_sed_bug. Vi shows the contents of this file (with :set list enabled) as:

don't$
delete$
me$
please$

I then run the following command:

sudo sed -n -i.bak /please/a\testing ./potential_sed_bug

Before we discuss the results; here is what the sed man page says:

-n, --quiet, --silent suppress automatic printing of pattern space

and

-i[SUFFIX], --in-place[=SUFFIX] edit files in place (makes backup if extension supplied). The default operation mode is to break symbolic and hard links. This can be changed with --follow-symlinks and --copy.

I've also looked other sed command references to learn how to append with sed. Based on my understanding from the research I've done; the resulting file content should be:

don't
delete
me
please
testing

However, running sudo cat ./potential_sed_bug gives me the following output:

testing

In light of this discrepancy, is my understanding of the command I ran incorrect or is there a bug with sed/the environment?

Answer 1

tl;dr

• Don't use -n with -i: unless you use explicit output commands in your sed script, nothing will be written to your file.

• Using -i produces no stdout (terminal) output, so there's nothing extra you need to do to make your command quiet.

---

By default, sed automatically prints the (possibly modified) input lines to whatever its output target is, whether implied or explicitly specified: by default, to stdout (the terminal, unless redirected); with -i, to a temporary file that ultimately replaces the input file.

In both cases, -n suppresses this automatic printing, so that - unless you use explicit output functions such as p or, in your case, a - nothing gets printed to stdout / written to the temporary file.

• Note that the automatic printing applies to the so-called pattern space, which is where the (possibly modified) input is held; explicit output functions such as p, a, i and c do not print to the pattern space (for potential subsequent modification), they print directly to the target stream / file, which is why a\testing was able to produce output, despite the use of -n.

Note that with -i, sed's implicit printing / explicit output commands only print to the temporary file, and not also to stdout, so a command using -i is invariably quiet with respect to stdout (terminal) output - there's nothing extra you need to do.

---

To give a concrete example (GNU sed syntax).

^{Since the use of -i is incidental to the question, I've omitted it for simplicity. Note that -i prints to a temporary file first, which, on completion, replaces the original. This comes with pitfalls, notably the potential destruction of symlinks; see the lower half of this answer of mine.}

# Print input (by default), and append literal 'testing' after 
# lines that contain 'please'.
$ sed '/please/ a testing' <<<$'yes\nplease\nmore'
yes
please
testing
more

# Adding `-n` suppresses the default printing, so only `testing` is printed.
# Note that the sequence of processing is exactly the same as without `-n`:
# If and when a line with 'please' is found, 'testing' is appended *at that time*.
$ sed -n '/please/ a testing' <<<$'yes\nplease\nmore'
testing

# Adding an unconditional `p` (print) call undoes the effect of `-n`.
$ sed -n 'p; /please/ a testing' <<<$'yes\nplease\nmore'
yes
please
testing
more