Another day, another bash and ssh discovery.

I’ve been working on a script which, in its most minimal state, does the following:

fun() {
  ssh -i key user@host 'true'
  return
}
echo before
fun
echo after

Assuming the key exists and true will indeed run on the foreign server, what will the output be when running this script? As it turns out, it depends on how you run it:

$ bash ./test.sh
before
after

$ cat ./test.sh | bash
before

Huh? Why isn’t bash executing the second echo?

In fact, it seems that the script just completely finishes after the fun() function returns. We see this when running the script in :

cat test.sh | bash -x
+ echo before
before
+ fun
+ ssh -i key user@host true
+ return

After some time reading the ssh manpage, I noticed:

-n      Redirects stdin from /dev/null (actually, prevents reading from stdin).  This must be used when ssh is run in the background.

Right. ssh by default reads from stdin unless disabled. Since the script is being read from stdin line by line, when the fun() function is actually executed, stdin still contains echo after, and as such, ssh reads it or effectively eats it.

---

There are a few flags you can use to disable the reading of stdin by default, like the aforementioned -n, but also -f. Alternatively, you can feed /dev/null into it, as such:

fun() {
  ssh -i key user@host 'true' < /dev/null
  return
}
echo before
fun
echo after