Bash script does not exit in loop

Question

I have run at below script in HP/UX and get that output:

Exiting #1
Exiting #2

But I expect that output:

Exiting #1

Script file:

data="aaa; bbb; ccc"

echo $data | while IFS=';' read -ra array; do
    echo "Exiting #1"
    exit -1
done


echo "Exiting #2"

exit 0

How can I solve this problem? Thanks.

Be aware that -1 is not a valid exit status for a process; it must be a non-negative value. — chepner
– chepner, Commented Dec 27, 2013 at 14:51

chepner · Accepted Answer · 2013-12-27 14:51:21Z

4

Because of the pipe, the command in the loop run in a subshell. When you call exit, only the subshell is terminated, and not the parent process.

You can overcome this by using shopt -s lastpipe, by using process substitution

while IFS=';' read -ra array; do
    echo "Exiting #1"
    exit -1
done < <(echo "$data")

or by using a here-string

while IFS=';' read -ra array; do
    echo "Exiting #1"
    exit -1
done <<< "$data"

chepner

538k77 gold badges594 silver badges746 bronze badges

answered Dec 27, 2013 at 10:35

user000001

33.7k14 gold badges86 silver badges113 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

P.P · Accepted Answer · 2013-12-27 10:35:33Z

2

When you do exit -1 inside the loop, it only exits the subprocess that was created due the pipe. Re-write it to use here-string:

 while IFS=';' read -ra array; do
    echo "Exiting #1"
    exit -1
done <<< ${data}

Now, you'll get the expected output.

answered Dec 27, 2013 at 10:35

P.P

122k20 gold badges193 silver badges251 bronze badges