Question

I’m trying to populate a shell variable called $recipient which should contain a value followed by a new-line.

$ set -x # force bash to show commands as it executes them

I start by populating $user, which is the value that I want to be followed by the newline.

$ user=[e-mail]
+ user=[e-mail]

I then call echo $user inside a double-quoted command substitution. The echo statement should create a newline after $user, and the double-quotes should preserve the newline.

$ recipient="$(echo $user)"
++ echo [e-mail]
+ recipient=[e-mail]

However when I print $recipient, I can see that the newline has been discarded.

$ echo "'recipient'"
+ echo ''\''recipient'\'''
'recipient'

I’ve found the same behaviour under bash versions 4.1.5 and 3.1.17, and also replicated the issue under dash.

I tried using “printf” rather than echo; this didn’t change anything.

Is this expected behaviour?

Answer 1

Command substitution removes trailing newlines. From the standard:

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 characters at the end of the substitution. Embedded characters before the end of the output shall not be removed; however, they may be treated as field delimiters and eliminated during field splitting, depending on the value of IFS and quoting that is in effect. If the output contains any null bytes, the behavior is unspecified.

You will have to explicitly add a newline. Perhaps:

recipient="$user
"

There’s really no reason to use a command substitution here. (Which is to say that $(echo ...) is almost always a silly thing to do.)