Bash script: How do I make the output of a c program store inside a variable as a column and not a row

Question

$ c=$(./getnumbers -n 5)
$ echo $c
1 2 3 4 5
$ ./getnumbers -n 5
1
2
3
4
5

How do I get it to store inside the variable as a column and not a row?

echo $c does not show what's in the variable. The lack of quotes may interfere, this apparently happens here. echo "$c" is better, printf '%s\n' "$c" even better. In general a variable may include characters or sequences that make the terminal do something (rather than print something), so you need printf '%s' "$c" | od -c or similar command to really tell what's in the variable; or (in Bash) declare -p c. — Kamil Maciorowski, Oct 02 '22 at 21:35
Does this answer your question? Why does my shell script choke on whitespace or other special characters? — Kamil Maciorowski, Oct 02 '22 at 21:35
Does this answer your question? Why do newline characters get lost when using command substitution? — G-Man Says 'Reinstate Monica', Oct 03 '22 at 08:49

score 1 · Answer 1 · answered Oct 03 '22 at 02:00

The output is stored "as a column", i.e. with the \n between lines preserved:

$ c=$(./getnumbers -n 5)
$ echo $c
1 2 3 4 5
$ echo "$c"
1
2
3
4
5

Without the " quotes, what is happening is that the shell is passing the output lines to echo as individual arguments. echo in turn then does its job and outputs them all - with only a space between them. With the " quotes, $c is passed to echo without being interpreted by the shell, so echo is then able to output the value correctly, as it was captured.

Bash script: How do I make the output of a c program store inside a variable as a column and not a row

1 Answers1