Bash while loop behave differently in file vs direct execution

Question

The following is the sample while loop.

res=()
lines=$'first\nsecond\nthird'
while read line; do
res+=("$line")
done <<< "$lines"
echo $res

When i run this directly in terminal im getting the following out put.

first second third

But when run the same script by saving it to a file. Then im getting the following out put.

first

Why is it behaving differently?

Note: I tested with and without shebang in file. result is same.

Look at the output of `declare -p lines`; in the case where you see all three entries, it's actually a single element containing linebreaks (which are removed by the unquoted parameter expansion). — Benjamin W., Nov 20 '20 at 16:43
[ShellCheck](https://www.shellcheck.net) automatically detects [this](https://github.com/koalaman/shellcheck/wiki/SC2128) and other common problems — that other guy, Nov 20 '20 at 16:54
I'm guessing when you run it "directly" your shell is not bash but zsh. Only printing `first` is correct/expected/normal behavior when you have `res=( first second third )` and then run `echo $res`; you would need to use `"${res[@]}"` to expand all three elements. — Charles Duffy, Nov 20 '20 at 16:55
@CharlesDuffy, yes, i forgot the fact that it is zsh. Thanks!!!. — lokanadham100, Nov 21 '20 at 13:51

score 0 · Answer 1 · answered Nov 20 '20 at 16:53

0

If you have a string with embedded newlines and want to turn it into an array split on them, use the mapfile builtin:

$ mapfile -t res <<<$'first\nsecond\nthird'
$ printf "%s\n" "${res[@]}"
first
second
third

answered Nov 20 '20 at 16:53

Shawn

1 Answers1