print newline in process substitution

Question

This is in bash(5.0.3(1)-release) / ubuntu 19

When I ran :

printf "%s" $'\n'

It printed out a new line.

Now when I run:

result="$(printf "%s" $'\n')"
printf "<%s>" "$result"

I expect $result contains a newline, but it's empty.

Can someone explain ?

Try it like this `result="$(printf "1%s" $'\n2')"; printf "<%s>" "$result"` — l'L'l, Jan 07 '20 at 11:39
I know we can do that, I just want an explanation of what happens in my post. — Philippe, Jan 07 '20 at 11:42

KamilCuk · Answer 1 · 2020-01-07T11:49:22.953

From posix shell manual, from chapter about process substitution, emphasis mine:

The shell shall expand the command substitution by executing command in a subshell environment (see Shell Execution Environment) and replacing the command substitution (the text of command plus the enclosing "$()" or backquotes) with the standard output of the command, removing sequences of one or more <newline>s at the end of the substitution.

It's impossible to store trailing newlines with process substitution. You can use for example printf -v bash extension:

printf -v result "%s" $'\n'

For most portability, I go with encoding the string in hex using od or xxd and store the string in hex. Or just store the resulting string in a file.

print newline in process substitution

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